# Differentiation with respect to a function

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1. May 8, 2015

### Gabriel Maia

Hi. A very long problem brought me to a derivative in the form

$\frac{\mathrm{d}f(x)}{\mathrm{d}g(x)}$

I'm assuming that

$\mathrm{d}g(x)=\left(\frac{\mathrm{d}g(x)}{\mathrm{d}x}\right)\mathrm{d}x$

So, is it correct to say that

$\frac{\mathrm{d}f(x)}{\mathrm{d}g(x)}=\left(\frac{\mathrm{d}g(x)}{\mathrm{d}x}\right)^{-1}\frac{\mathrm{d}f(x)}{\mathrm{d}x}$?

Thank you

2. May 8, 2015

### certainly

Yes.
(P.S. I don't think "derivating" is a word..............)

3. May 8, 2015

### Gabriel Maia

differentiation, perhaps?

Thank you very much.

4. May 8, 2015

### certainly

It is interesting to note that this does not work with higher derivatives........
[EDIT:- so $d^2g(x)\neq \Big(\frac{d^2g(x)}{dx^2}\Big) dx^2$]

Last edited: May 8, 2015
5. May 8, 2015

### Ray Vickson

We have
$$\frac{d\, f(x)}{d\, g(x)} = \lim_{\Delta g(x) \to 0} \frac{ \Delta f(x)}{\Delta g(x)},$$
where
$$\Delta f(x) = f(x + \Delta x) - f(x) \doteq f'(x) \Delta x, \\ \Delta g(x) = g(x + \Delta x) - g(x) \doteq g'(x) \Delta x,$$
hence
$$\frac{d f(x)} {d g(x)} = \frac{f'(x)}{g'(x)}$$
This is equivalent to what you wrote.