How do you differentiate #f(x)= 1/(e^(3x) -x)# using the quotient rule?

Question

How do you differentiate #f(x)=  1/(e^(3x)  -x)# using the quotient rule?

Emma Aldridge · Accepted Answer

The derivative is #-(3e^(3x)-1)/[e^(3x)-x]^2#.

Here's the quotient rule: #d/dx(f(x)/g(x))quad=quad(f'(x)g(x)-f(x)g'(x))/[g(x)]^2# Let's use this in our problem: #color(white)=d/dx(1/(e^(3x)-x))# #=(d/dx(1)*(e^(3x)-x)-1*d/dx(e^(3x)-x))/[e^(3x)-x]^2# #=(color(red)cancelcolor(black)(0*(e^(3x)-x))-1*d/dx(e^(3x)-x))/[e^(3x)-x]^2# #=(-1*d/dx(e^(3x)-x))/[e^(3x)-x]^2# #=-(d/dx(e^(3x))-d/dx(x))/[e^(3x)-x]^2# #=-(3e^(3x)-d/dx(x))/[e^(3x)-x]^2# #=-(3e^(3x)-1)/[e^(3x)-x]^2# That's it. Hope this helped!

Paisley Johnson · Answer

undefined To differentiate $ f(x) = \frac{1}{e^{3x} - x} $ using the quotient rule, you follow this formula:

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

Where $ u = 1 $ and $ v = e^{3x} - x $.

Therefore, $ u' = 0 $ and $ v' = 3e^{3x} - 1 $.

Plugging into the quotient rule formula, you get:

$$ \frac{d}{dx} \left( \frac{1}{e^{3x} - x} \right) = \frac{(0)(e^{3x} - x) - (1)(3e^{3x} - 1)}{(e^{3x} - x)^2} $$

Simplify the expression:

$$ \frac{d}{dx} \left( \frac{1}{e^{3x} - x} \right) = \frac{-3e^{3x} + 1}{(e^{3x} - x)^2} $$

Elijah Abram · Answer

undefined To differentiate the function $ f(x) = \frac{1}{e^{3x} - x} $ using the quotient rule, we apply the formula:

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

Where:
- $ u = 1 $
- $ u' = 0 $ (since the derivative of a constant is zero)
- $ v = e^{3x} - x $
- $ v' = (e^{3x})' - (x)' $

Now, let's compute the derivatives:
- $ u' = 0 $
- $ v' = 3e^{3x} - 1 $

Substitute these into the quotient rule formula:

$$ \frac{d}{dx} \left( \frac{1}{e^{3x} - x} \right) = \frac{(0)(e^{3x} - x) - (1)(3e^{3x} - 1)}{(e^{3x} - x)^2} $$

Simplify:

$$ = \frac{-3e^{3x} + 1}{(e^{3x} - x)^2} $$

Thus, the derivative of $ f(x) = \frac{1}{e^{3x} - x} $ using the quotient rule is $ \frac{-3e^{3x} + 1}{(e^{3x} - x)^2} $.