# Partial Derivatives

1. Jul 1, 2007

### p4nda

Find the four second partial derivatives for f(x,y) = y^2e^x + xycosx

I am stuck on the last part... here's what I got so far:

Zx = y^2e^x - ysinx
Zxx = y^2e^x - ycosx
Zxy = ?

Zy = 2ye^x + xcosx
Zyy = 2e^x + xcosx
Zyx = ?

I need help with solving for xy. Both should end up being the same.

2. Jul 1, 2007

### mjsd

the trick is to ignore the other variable whenever you are not differentiating with respect to it. eg.
$$\frac{\partial}{\partial x} x e^y = e^y$$ and
$$\frac{\partial}{\partial y} x e^y = x e^y$$

in first case you treat y as if it is a constant whereas in the second you treat x as constant instead. the same rules apply when you go to higher derviatives
eg.
$$\frac{\partial}{\partial x} \frac{\partial}{\partial y} (x e^y) = \frac{\partial}{\partial x}(x e^y) = e^y$$

3. Jul 1, 2007

### mjsd

by the way your Zx is wrong