Differentiation with respect to a function

[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

 

[certainly]

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

 

[Gabriel Maia]

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

differentiation, perhaps?

Thank you very much.

 

[certainly]

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

 

[Ray Vickson]

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

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&#039;(x) \Delta x, \\<br /> \Delta g(x) = g(x + \Delta x) - g(x) \doteq g&#039;(x) \Delta x,
hence
\frac{d f(x)} {d g(x)} = \frac{f&#039;(x)}{g&#039;(x)}
This is equivalent to what you wrote.