# Homework Help: Partial Differentiation (very simple mistake, I think!)

1. Apr 15, 2010

### henryc09

1. The problem statement, all variables and given/known data
[PLAIN]http://img408.imageshack.us/img408/7163/partialdifferent.jpg [Broken]

So this means differentiate w.r.t y first, so I want dz/dy, and then w.r.t x right?

so I rearrange so that y=z3/3 + xz
and differentiate w.r.t z to get dy/dz, and then do 1 over this which i get as:

1/(z2+x)

and then differentiating this w.r.t x, I get
-1/(z2+x)2

which isn't quite right. Can anyone see where I've gone wrong? Thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 4, 2017
2. Apr 15, 2010

### vela

Staff Emeritus
You need to use the chain rule. Don't forget z still varies with x.

3. Apr 15, 2010

### rs1n

It may help to rewrite $z$ as $z(x,y)$ -- looks clunkier, but I find that it helps me remember that I have multivariate function.