# Simple question about partial derivatives

1. Aug 21, 2011

### AlexChandler

I have come to a bit of a misunderstanding with partial derivatives. I will try to illustrate my problem. Say we have a function f(x, y(x), y'(x)) where y'(x)=dy/dx. Now suppose that f does not explicitly depend on x. My physics book says at this point that ∂f/∂x=0, even though y(x) and y'(x) may depend on x.
Suppose f=y2
Then ∂f/∂x=0
but if we have y(x)=x, then we can write f as:
f=x2 and we have
∂f/∂x=2x
How can we have two different answers for the same derivative by simply rewriting the function in a different way?
I apologize in advance if the answer is obvious and I am being a bit annoying by asking. But if you do have a helpful comment to post, I would greatly appreciate it!
-Alex

2. Aug 21, 2011

### mathman

In the first case, ∂f/∂y = 2y. Note that (I'll ignore y' since it is irrelevant):
df/dx = ∂f/∂x + (∂f/∂y)(dy/dx). Therefore the final result is the same in both cases.

3. Aug 22, 2011

### HallsofIvy

The partial derivative of f "with respect to y" assumes y is independent of x or any other variables. If, in fact, f is a function of both x and y, and y is a function of x, then
$$\frac{df}{dx}= \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx}$$

In your example, with $f(x,y)= y^2$ and y= x,
$$\frac{\partial f}{\partial x}= 0$$
$$\frac{\partial f}{\partial y}= 2y$$
and
$$\frac{df}{dx}= 0+ (2y)(1)= 2x$$.