Question

How do you differentiate `f(x)= (x-e^(x))/(2-3x)` using the quotient rule?

Answer 1

`f'(x) = (e^x(3x-5) + 2)/(2-3x)^2`

Explanation:

The quotient rule is as follows:

`d/dx[color(blue)(f(x))/color(red)(g(x))] = (color(blue)(f'(x))*color(red)(g(x)) - color(red)(g'(x))*color(blue)(f(x)))/[color(red)g(x)]^2`

We know that for this problem, `f(x) = (x-e^(x))`, and `g(x) = (2-3x)` So, we can just plug these in:

`d/dx[color(blue)((x-e^(x)))/color(red)((2-3x))] = (color(blue)(d/dx(x-e^(x)))*color(red)((2-3x)) - color(red)(d/dx(2-3x))*color(blue)((x-e^(x))))/[color(red)(2-3x)]^2`

This is actually the hardest part of this problem. The rest is just taking a bunch of basic derivatives. I'll leave that process to you, so the final answer you should end up with is:

`((1-e^x)(2-3x) + 3(x-e^x))/(2-3x)^2`

You'd usually just leave it like this, but for this particular example you can simplify it a bit further. If you foil out all the terms in the numerator, you eventually get:

`(e^x(3x-5) + 2)/(2-3x)^2`

Which looks nicer.

If you'd like additional help on the concepts associated with this problem, I have some videos that can help:
Product & Quotient Rules
Practice Problem Video (contains some extra practice problems)

Additionally, here's my answer to a slightly more complicated quotient rule question, right here on socratic. Hopefully it can give you some context for how this principle is applied across different kinds of problems.

Hope that helped :)

Answer 2

`f'(x)=(3xe^x-5e^x+2)/(2-3x)^2`

Explanation:

`"given "f(x)=(g(x))/(h(x))" then"`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr"quotient rule"`

`g(x)=x-e^xrArrg'(x)=1-e^x`

`h(x)=2-3xrArrh'(x)=-3`

`rArrf'(x)=((2-3x)(1-e^x)-(x-e^x)(-3))/(2-3x)^2`

`color(white)(rArrf'(x))=(2-2e^xcancel(-3x)+3xe^xcancel(+3x)-3e^x)/(2-3x)^2`

`color(white)(rArrf'(x))=(3xe^x-5e^x+2)/(2-3x)^2`