# Derivative w.r.t. a function

1. Aug 2, 2007

### ledol83

Hi, i have a question on taking derivative w.r.t. to a function (instead of an independent variable). Actually i saw an excellent post on this same forum but that one was about a single variable.

My question is: f is function of x and y, and z is some other function also dependent on x and y, so is the following correct?

df/dz=df/dx*dx/dz+df/dy*dy/dz

it differs from the classic chain rule in the sense that z is actually a function (not an independent var), so i am not sure about this.

I appreciate so much for any comment!

2. Aug 3, 2007

### lalbatros

First, go back to how the derivative wrt to a function is defined. (functional derivative)
As I understood, f is a function of x and y: f(x,y)
therefore, if you confirm that, I would say the df/dz = 0 .

3. Aug 3, 2007

### ledol83

Hi actually i have f(x,y) and z(x,y) and was just wondering if this is true:

df/dz=df/dx*dx/dz+df/dy*dy/dz

thanks a lot!!

4. Aug 3, 2007

### lalbatros

No this cannot be true, since this has no meaning.
Tell us what you think the meaning of df/dz would be, maybe then we can help.

5. Aug 3, 2007

### ledol83

i have realized that what i posed was not meaningful. i am now thinking over my problem again.. thanks!

6. Aug 3, 2007

### ice109

how could you have a derivative of a function with respect to another function that the first function is not a function of :rofl::rofl:

7. Aug 3, 2007

### HallsofIvy

If f(x) is a function of x and g(x) is a function of x, you can surely write f as a function of g.

In particular,
$$\frac{df}{dg}= \frac{df}{dx}\frac{dx}{dg}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}$$

8. Aug 3, 2007

### ice109

not that i'm doubting you personally but i don't see where this comes from; proof?

9. Aug 5, 2007

### lalbatros

"not that i'm doubting you personally but i don't see where this comes from; proof?"

The first point is to know when we are talking about f1=f(g) or f2=f(x),
The second point is about the class of functions considered,
Otherwise, this is trivial (assuming dg is smooth):

f'(g) = f(g+dg)/dg = f(g(x)+dg(x))/dg(x) = f(g(x) + g'(x) dx) / (g'(x) dx) = f'(x)/g'(x)

10. Aug 5, 2007

### AiRAVATA

It is a consequence of the chain rule and the inverse function theorem. Maybe you should do some reading too instead of making fun of peoples questions. Better try to help or keep out.

Last edited: Aug 5, 2007