# Partial Derivatives

• Oct 22nd 2011, 02:31 PM
freestar
Partial Derivatives
Calculate $\displaystyle x(\frac{\partial z}{\partial x})-y(\frac{\partial z}{\partial y})$ for $\displaystyle z=f(xy)$ where f is some differentiable function.

I am not sure if I am doing it right. This is what I have tried so far.

$\displaystyle \frac{\partial z}{\partial x} = \frac{\partial f}{\partial x}(y)$

$\displaystyle \frac{\partial z}{\partial y} = \frac{\partial f}{\partial y}(x)$

• Oct 22nd 2011, 05:37 PM
Ackbeet
Re: Partial Derivatives
I find it always helpful to think of a function as a function of a certain number of independent variables. In this case, f is a function of one variable. I know it's got an x and a y in there, but if it were a function of two variables, I think you'd write as f(x,y) = g(xy), and even then, g would be a function of only one variable. How can that be? Because you could write a new variable u = xy, and then say that z = f(u).

Hence, I think, rather, that

$\displaystyle \frac{\partial z}{\partial x}=y\,f'(xy),$ and so on. Does that make sense?
• Oct 22nd 2011, 07:54 PM
freestar
Re: Partial Derivatives
Ahh makes sense. So the answer would this be the right answer?:
$\displaystyle xyf'(xy)-xyf'(xy)$ which would be equal to 0?
• Oct 24th 2011, 02:15 AM
Ackbeet
Re: Partial Derivatives
Quote:

Originally Posted by freestar
Ahh makes sense. So the answer would this be the right answer?:
$\displaystyle xyf'(xy)-xyf'(xy)$ which would be equal to 0?

Yep, that's what I get.