1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentiation with respect to a function

  1. May 8, 2015 #1
    Hi. A very long problem brought me to a derivative in the form

    [itex]\frac{\mathrm{d}f(x)}{\mathrm{d}g(x)}[/itex]


    I'm assuming that

    [itex]\mathrm{d}g(x)=\left(\frac{\mathrm{d}g(x)}{\mathrm{d}x}\right)\mathrm{d}x[/itex]

    So, is it correct to say that

    [itex]\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}[/itex]?

    Thank you
     
  2. jcsd
  3. May 8, 2015 #2
    Yes.
    (P.S. I don't think "derivating" is a word..............)
     
  4. May 8, 2015 #3
    differentiation, perhaps?

    Thank you very much.
     
  5. May 8, 2015 #4
    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##]
     
    Last edited: May 8, 2015
  6. May 8, 2015 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    We have
    [tex] \frac{d\, f(x)}{d\, g(x)} = \lim_{\Delta g(x) \to 0} \frac{ \Delta f(x)}{\Delta g(x)}, [/tex]
    where
    [tex] \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, [/tex]
    hence
    [tex] \frac{d f(x)} {d g(x)} = \frac{f'(x)}{g'(x)} [/tex]
    This is equivalent to what you wrote.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted