chain rule

**chris25** · July 7th 2008, 11:04 AM

use the appropriate Chain rule to find the partial derivate of z with respect to u
for z=(cos x) (sin y), x=u-v , y=u^2+v^2
write your answer as a function of u and v.

I dont understand how to write it in terms of u and v and i think i am messing up the partial derivatives can any one help me???

**TheEmptySet** · July 7th 2008, 12:58 PM

Originally Posted by chris25

use the appropriate Chain rule to find the partial derivate of z with respect to u
for z=(cos x) (sin y), x=u-v , y=u^2+v^2
write your answer as a function of u and v.

I dont understand how to write it in terms of u and v and i think i am messing up the partial derivatives can any one help me???

$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u} =-\sin(x)\sin(y)(1)+\cos(x)\cos(y)(2u)$

From here we know what x and y are as functions of u and v so just plug them in.

$\frac{\partial z}{\partial u}=\sin(u-v)\sin(u^2+v^2)(1)+\cos(u-v)\cos(u^2+v^2)(2u)$

Good luck.

**jme44** · July 9th 2008, 01:03 PM

then you need to do the same with respect to v right?

**Mathstud28** · July 9th 2008, 05:16 PM

Originally Posted by jme44

then you need to do the same with respect to v right?

Yes, you can. Just apply the same concept.

If $f\equiv{f(x,y)}$

where

$x=x(u,v)$

and

$y=y(u,v)$

Then,

$\frac{\partial{f}}{\partial{v}}=\frac{\partial{f}} {\partial{x}}\cdot\frac{\partial{x}}{\partial{v}}+ \frac{\partial{f}}{\partial{y}}\cdot\frac{\partial {y}}{\partial{v}}$

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