Question

f(x) = x/(x +1)
Find g'(x), where g(x) is the inverse of f(x)

This is what I've done so far:

f'(x) = 1/(x+1)^2 by quotient rule
g(x) = -x/(x-1)

Using derivative of the inverse theorem, g'(x) = 1/[f'(g(x))], I get:

1/[(-x/(x-1)) +1]^2

The homework software tells me this is incorrect. I've been going over this problem for more than an hour and I'm going crazy trying to figure out where my error is. Please help!

Answer 1

agree that g(x)=x/(1-x)
now g'(x)=1/(1-x)^2

wrong or correct?

Answer 2

Correct, but why didn't I get the right answer using the theorem?

Answer 3

Or did I just not simplify it enough?

Answer 4

don't know...
i'm too lazy :)
I took a shortcut & just took derivative of g(x) that you identified correctly (by the way)...

Answer 5

That's what I did to verify your answer. lol

Answer 6

may yet may be no... why complicate?

Answer 7

Because that's what the instructions told me to use to solve the problem.

Answer 8

Oups... I didn't see that...
let me look now

Answer 9

i think I got it!
here we go:

Answer 10

f'=1/(x+1)^2
agreed?
\[g'(x)=1/1/(x+1)^{2}=(x+1)^{2}=(-x/(x-1) +1)^{2}=
\[=[(-x+x-1)/(x-1)]^{2}=1/(x-1)^{2}\]

Answer 11

shoot... it's cut it..
\[=[(-x+x-1)/(x-1)]^{2}=1/(x-1)^{2}\]

Answer 12

do you agree?

Answer 13

Simple error. Your inverse is identified correctly. The derivative of your original function is also correct. Plugging them into the inverse derivative theorem gives:
\[g \prime=1/[(1/-x/(x-1))+1)]^2=1/[((x/(1-x))+1)]^2\]
This gives,\[=[(1-x+x)/x]^2=x^2\]Your final answer is x^2

Answer 14

disagree... the checked the answer before.
the correct on is 1/(x-1)^2
we just tried different ways to solve it

Answer 15

yeah I just noticed that let me check my arithmetic

Answer 16

Yeah my arithmetic is wrong:\[=1/[1/((x/(1-x)+1)]^2\]
\[=1/[1/((x+1-x)/(1-x))]^2\]
\[=1/(1/(1/1-x))^2\]
=\[1/(1-x)^2\]
So many brackets! Sorry if I screwed any up. Your essential error is you plugged the stuff into the formula wrong.