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