
Derivative Problem
Would someone tell me how would you solve the following:
1) Find $\displaystyle \frac{\partial z}{\partial y}$
$\displaystyle z=sin(x^3 e^{y}4y^5)$
this is what i have done, but i don't think its correct.
$\displaystyle \frac{\partial z}{\partial y}=x^3 e^y20y^4cos(x^3 e^{y}4y^5)$
P.S

$\displaystyle z = \sin{(x^3e^y  4y^5)}$.
Let $\displaystyle u = x^3e^y  4y^5$ so that $\displaystyle z = \sin{u}$.
$\displaystyle \frac{\partial u}{\partial y} = x^3e^y  20y^4$.
$\displaystyle \frac{dz}{du} = \cos{u} = \cos{(x^3e^y  4y^5)}$.
Therefore $\displaystyle \frac{\partial z}{\partial y} = (x^3e^y  20y^4)\cos{(x^3e^y  4y^5)}$.
I agree with your answer, as long as you put the brackets where they're needed.