Derivative Problem

**Paymemoney** · September 28th 2010, 08:16 PM

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

**Prove It** · September 28th 2010, 08:37 PM

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

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.

Thread: Derivative Problem

LinkBack

Thread Tools

Display

Derivative Problem

Similar Math Help Forum Discussions

Derivative Problem

problem with derivative

derivative problem -

Another derivative problem

another derivative problem

Search Tags