How do you differentiate #f(x)= (x+5)/ (x- 7 )# using the quotient rule?

Question

How do you differentiate #f(x)=  (x+5)/  (x- 7 )# using the quotient rule?

Elijah Abram · Accepted Answer

#f'(x)=-12/(x-7)^2# Given: #f(x)=(N(x))/(D(x))# #f'(x)=(N'(x)*D(x)-D'(x)*N(x))/((D(x))^2)# Then: #f'(x)=(1*(x-7)-1*(x+5))/(x-7)^2=(cancel(x)-7cancel(-x)-5)/(x-7)^2=# #=-12/(x-7)^2#

Christopher Agresta · Answer

undefined To differentiate $ f(x) = \frac{x + 5}{x - 7} $ using the quotient rule:

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

Where:
- $ u = x + 5 $
- $ v = x - 7 $

Now, find the derivatives $ u' $ and $ v' $:

$$ u' = \frac{d}{dx}(x + 5) = 1 $$
$$ v' = \frac{d}{dx}(x - 7) = 1 $$

Substitute the values into the quotient rule formula:

$$ f'(x) = \frac{(1)(x - 7) - (x + 5)(1)}{(x - 7)^2} $$
$$ = \frac{x - 7 - x - 5}{(x - 7)^2} $$
$$ = \frac{-12}{(x - 7)^2} $$

So, the derivative of $ f(x) $ with respect to $ x $ using the quotient rule is:

$$ f'(x) = \frac{-12}{(x - 7)^2} $$

Christopher Agresta · Answer

undefined To differentiate the function $ f(x) = \frac{x+5}{x-7} $ using the quotient rule, you apply the formula:

$$ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} $$

Where $ u = x + 5 $ and $ v = x - 7 $. Differentiating $ u $ and $ v $ with respect to $ x $ gives:

$$ \frac{du}{dx} = 1 $$
$$ \frac{dv}{dx} = 1 $$

Now, applying the quotient rule:

$$ f'(x) = \frac{(x - 7)(1) - (x + 5)(1)}{(x - 7)^2} $$

Simplify the numerator:

$$ f'(x) = \frac{x - 7 - x - 5}{(x - 7)^2} $$

$$ f'(x) = \frac{-12}{(x - 7)^2} $$

So, the derivative of $ f(x) $ with respect to $ x $ is $ \frac{-12}{(x - 7)^2} $.