# Partial derivative concept

1. Apr 27, 2008

### fk378

1. The problem statement, all variables and given/known data
Given the partial derivative df/dx= 3-3(x^2)

what is d^2f/dydx?

I'm not sure if the answer would be 0, since x is held constant, or if it would remain 3-3(x^2) (since df/dx is a function of x now?)

2. Apr 27, 2008

### G01

The answer is one of those choices. Here, think about it like this:

You are given a function:

$$g(x)=3-3x^2$$

You want to find: $$\frac{\partial g}{\partial y}$$

What is that derivative? Now, what if: $$g(x)=\frac{\partial f}{\partial x}$$

Does this change the partial derivative of g with respect to y?

3. Apr 27, 2008

### Dick

You were right the first time. With x held constant the d/dy is just differentiating a constant. It's 0.

4. Apr 27, 2008

### HallsofIvy

Staff Emeritus
As is always true, with "nice" functions, the two mixed derivatives are equal. You could find $\partial^2 f/\partial x\partial y$ by differentiating first with respect to x, then with respect to y: first getting -6x and then, since it does not depend on y, 0. Or you could differentiate first with respect to y, then with respect to x: getting 0 immediately and then, of course, the derivative of "0" with respect o x is 0.

I, and I suspect many who read your post, was momentarily taken aback since I thought you were "holding x constant" through both derivatives. But you are correct: since this function does not depend on y, any derivative of it with respect to y, is 0.