How do you differentiate #f(x)=4/sqrt(x-2)# using the quotient rule?

Question

Emma Aldridge · Accepted Answer

#f'(x)=(-2)/(x-2)^(3/2)#

Quotient rule states that

#d/dx[f(x)/(g(x))]=(g(x)*f'(x)-f(x)*g'(x))/([g(x)]^2#.

So in this case:

#d/dx[4]=0#

#d/dx[sqrt(x-2)]=1/2(x-2)^(-1/2)*(1)#

#f'(x)=(sqrt(x-2) * (0)-4(1/2)(x-2)^(-1/2) * (1))/(x-2)#

#=(-2(x-2)^(-1/2))/(x-2)#

#=(-2)/(x-2)^(3/2)#

Emily Ammerman · Answer

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The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

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Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

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Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

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The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

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Now, find the derivatives of $ u(x) $ and $ vTo differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

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Now, find the derivatives of $ u(x) $ and $ v(x)To differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

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Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

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Now, find the derivatives of $ u(x) $ and $ v(x) $:

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Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ uTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)vTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)vTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

NowTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now,To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, findTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrtTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find theTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivativesTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives ofTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $To differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(xTo differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x)To differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) \To differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

SubTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ andTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

SubstituteTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $To differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these intoTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ vTo differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into theTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x)To differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient ruleTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ uTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ fTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x)To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x)To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) =To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) =To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \fracTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrtTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdotTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

SubTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

SubstituteTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrtTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute theseTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into theTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient ruleTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formulaTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ fTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x)To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) =To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \fracTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdotTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

STo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

SimplifyTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrtTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x)To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \fracTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdotTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\fracTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrtTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrtTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrtTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ fTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x)To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) =To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

STo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ fTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrtTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x)To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \fracTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \fracTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

ThisTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is theTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative ofTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ fTo differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(xTo differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x -To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x)To differentiate \( f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x - To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x) $To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x - 2To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x) $ usingTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x - 2}To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x) $ using theTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x - 2} \To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x) $ using the quotientTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x - 2} $$
$$To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x) $ using the quotient ruleTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x - 2} $$
$$ fTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x) $ using the quotient rule.To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x - 2} $$
$$ f'(To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x) $ using the quotient rule.To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x - 2} $$
$$ f'(xTo differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule, where $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{-\frac{4}{2\sqrt{x - 2}}}{x - 2} $$

$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x) $ using the quotient rule.To differentiate $ f(x) = \frac{4}{\sqrt{x - 2}} $ using the quotient rule:

Let $ u(x) = 4 $ and $ v(x) = \sqrt{x - 2} $.

The quotient rule states that the derivative of $ \frac{u(x)}{v(x)} $ is:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = 0 $$
$$ v'(x) = \frac{1}{2\sqrt{x - 2}} $$

Substitute these into the quotient rule formula:

$$ f'(x) = \frac{0 \cdot \sqrt{x - 2} - 4 \cdot \frac{1}{2\sqrt{x - 2}}}{(\sqrt{x - 2})^2} $$

Simplify:

$$ f'(x) = \frac{- \frac{4}{2\sqrt{x - 2}}}{x - 2} $$
$$ f'(x) = \frac{-2}{\sqrt{x - 2}(x - 2)} $$

This is the derivative of $ f(x) $ using the quotient rule.