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There's a question I need to solve, which requires that I take the derivative of some function by the square of a variable, and I'm not sure how to do such a thing.

For example: $\frac{dx}{d(x^2)}$ - I've tried $t=x^2$, $d\sqrt{t}/dt$ is easy enough to calculate, but is it the right way? $\sqrt{x^2}\ne x$ if $x<0$

So what's the right way to do it?

edit: I got my answer, but I forgot to ask: how would this look when using the definition of the derivative? I know that:

$df(x)/dx = lim_{\Delta \to 0}\frac{f(x+\Delta)-f(x)}{\Delta} $

But if you replace $\Delta$ with $\Delta^2$, wouldn't it be the same?

Vibhav
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user1999728
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  • Can you write out the full problem you are trying to solve? – fromGiants Jun 22 '14 at 17:05
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    Are you sure you aren't supposed to calculate the second derivative, i.e. $\mathrm{d}^2f(x) / \mathrm{d}x^2$? – leftaroundabout Jun 22 '14 at 17:06
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    The question is about taking the derivative of a certain function with respect to $x$ or $x^2$. @user2524719: I'd like to know how to do this in general, not just specifically for my problem... but my main problem is taking the derivative of $ln x$ w.r.t. $x^2$ – user1999728 Jun 22 '14 at 17:08

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$$\frac{df(x)}{d(x^2)}=\frac{df(x)}{dx}\frac{dx}{d(x^2)}=\frac{df(x)}{dx}\left(\frac{d(x^2)}{dx}\right)^{-1}=\frac{1}{2x}\frac{df(x)}{dx}$$

mike
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  • A friend of mine suggested this, wasn't sure it was actually a valid thing to do. Thank you :) – user1999728 Jun 22 '14 at 17:13
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    Thanks for accepting the answer and the upvote. I sometimes write $d(x^a)=ax^{a-1}dx$ in $\frac{df(x)}{d(x^a)}$ with $(a>0)$. – mike Jun 23 '14 at 00:16