# Thread:

1. ## Derivative Problem

Would someone tell me how would you solve the following:
1) Find $\frac{\partial z}{\partial y}$
$z=sin(x^3 e^{y}-4y^5)$

this is what i have done, but i don't think its correct.

$\frac{\partial z}{\partial y}=x^3 e^y-20y^4cos(x^3 e^{y}-4y^5)$

P.S

2. $z = \sin{(x^3e^y - 4y^5)}$.

Let $u = x^3e^y - 4y^5$ so that $z = \sin{u}$.

$\frac{\partial u}{\partial y} = x^3e^y - 20y^4$.

$\frac{dz}{du} = \cos{u} = \cos{(x^3e^y - 4y^5)}$.

Therefore $\frac{\partial z}{\partial y} = (x^3e^y - 20y^4)\cos{(x^3e^y - 4y^5)}$.

I agree with your answer, as long as you put the brackets where they're needed.
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