# Chain Rule Question.

• January 31st 2013, 04:13 PM
KhanDisciple
Chain Rule Question.
Find $\frac{dy}{dx}$ given that $y=\frac{u-17}{u+17}$ , $u=x^3$.

I began trying to solve it like this: $\frac{dy}{du}$ = $\frac{(u+17)(-17)-(u-17)(17)}{(u+17)^2}$ = $\frac{-34u}{(u+17)^2}$

$\frac{du}{dx}$ = $3x^2$

$\frac{dy}{dx}$ = $\frac{-34x^3}{(x^3+17)^2} (3x^2)$ = $\frac{-102x^5}{(x^3+17)^2}$

Can someone tell me what I am doing wrong?
Thanks
• January 31st 2013, 04:25 PM
emakarov
Re: Chain Rule Question.
$(u+17)'=1$, not 17.
• January 31st 2013, 10:12 PM
KhanDisciple
Re: Chain Rule Question.
Thank you emakarov, sometimes I think I am blind.