# Chain rule problem

• Feb 21st 2010, 07:09 AM
Kakariki
Chain rule problem
Hey, I have a question that I do not know how to go about answering.

Question:
Let $y = g(h(x))$where $h(x) = \frac{x^2}{x+2}$. If $g'(9/5) = -2$, find dy/dx when $x = 3$

Basically I have found the derivative of $h(x) = \frac{x^2}{x+2}$. Which is $h'(x) = \frac {-x^2}{(x + 2)^2} + \frac {2x}{(x+2)}$

I do not know what I am supposed to do with $g'(9/5) = -2$. How do I find what the function g(x) is from that?

Thanks!
• Feb 21st 2010, 07:13 AM
Ted
Quote:

Originally Posted by Kakariki
Hey, I have a question that I do not know how to go about answering.

Question:
Let $y = g(h(x))$where $h(x) = \frac{x^2}{x+2}$. If $g'(9/5) = -2$, find dy/dx when $x = 3$

Basically I have found the derivative of $h(x) = \frac{x^2}{x+2}$. Which is $h'(x) = \frac {-x^2}{(x + 2)^2} + \frac {2x}{(x+2)}$

I do not know what I am supposed to do with $g'(9/5) = -2$. How do I find what the function g(x) is from that?

Thanks!

Start by this:
If $y=g\left(h(x)\right)$ then $y'=g'\left(h(x)\right) h'(x)$.
• Feb 21st 2010, 07:54 PM
Kakariki
Thanks, that's the chain rule. I have figured out that g'(9/5) basically is the same thing as $\frac {(3)^2}{(3) + 2)}$ (subbing 3 in for x). What does this tell me? I still do not understand how to find what g(x) is.

Basically asking for a more detailed explanation.

Thanks for the response though!
• Feb 22nd 2010, 12:54 AM
HallsofIvy
Quote:

Originally Posted by Kakariki
Thanks, that's the chain rule. I have figured out that g'(9/5) basically is the same thing as $\frac {(3)^2}{(3) + 2)}$ (subbing 3 in for x). What does this tell me? I still do not understand how to find what g(x) is.

It tells you the answer! The problem says NOTHING about finding g(x) and there is no reason to do that. f(3)= 9/5 so the derivative of g(f(x)), at x= 3 is g'(9/5)f'(3).

Quote:

Basically asking for a more detailed explanation.

Thanks for the response though!