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Let $r,\theta \in \mathbb{R}$. As stated in the title, how do I differentiate $f(r\cos\theta) = r$ with respect to $r\cos\theta$? I have never encountered a question or concept like this, and am not sure where to start.

My first thought is to start from the fundamentals: perhaps I should try differentiating $x$ with respect to $2x$. Perhaps I can use change of variables with $u$ = $2x$. Then the problem would be equivalent to differentiating $\frac{x}{2}$ with respect to $x$, which is easy. However, evaluating either $\frac{d}{d(2x)} x$ or $\frac{d}{d(r\cos\theta)} r$ in various software produces error messages, so I'm not sure that the change of variables idea is even valid.

How should I proceed? Any advice is deeply appreciated.

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    $\begingroup$ I am guessing but what about replacing $u\rightarrow r\cos\theta$? $\endgroup$ Jul 2, 2020 at 19:35
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    $\begingroup$ Note that the typographical difference between $r cos \theta$ and $r\cos\theta$ is not only that in the latter $\cos$ is not italicized but also in the spacing, and the difference is in just one keystroke in the code, as in my edits to this question. $\endgroup$ Jul 2, 2020 at 19:37
  • $\begingroup$ I think some context is missing. My guess is that you have a function of $r$ and $\theta$ and you want to write the derivative with respect to $x$ (polar to Cartesian transform) $\endgroup$
    – Andrei
    Jul 2, 2020 at 19:38
  • $\begingroup$ @SamuelA.Morales That was what I was thinking, but I wasn't sure if change of variables is a valid approach. For example, in my "differentiate $x$ with respect to $2x$" toy example, do you think my change of variable strategy is valid? $\endgroup$
    – pythonuser
    Jul 2, 2020 at 19:39
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    $\begingroup$ I looks similar to this question math.stackexchange.com/q/3581852/399263 $\endgroup$
    – zwim
    Jul 2, 2020 at 19:41

2 Answers 2

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This depends on the dependence between $r$ and $\theta$. We can see that the differential

$$d(r\cos\theta)=\cos\theta dr-r\sin\theta d\theta$$

and so $$f'(r\cos\theta)=\frac{1}{\cos\theta-r\sin\theta\frac{d\theta}{dr}}.$$ Outside of this fact, the context is extremely important. Are you trying to solve for $f$? If so, there are many solutions. With a bit of practice, one can master turning functional equations into relatively monstrous differential equations (which is a general enough and delicate enough process that I leave it to you), and from there you can proceed to find solutions to the function.

If you suppose $u=r\cos\theta$, you'll end up in the same situation roughly. You could also assume a parametric surface, for example $S^1$. The context and intent here are pretty important.

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In $f(r\cos\theta)=r$, you can write $u=r\cos\theta$. Then $r=\frac u{\cos\theta}$. Finally $$\frac{df(r\cos\theta)}{d(r\cos\theta)}=\frac{df(u)}{du}=\frac 1{\cos\theta}$$

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  • $\begingroup$ @zwim Not clear from the question. What if $\theta$ is just a parameter and $f(x)=\frac{x}{\cos\theta}$ ? $\endgroup$
    – Andrei
    Jul 2, 2020 at 19:52
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    $\begingroup$ I think, you should explicitly state that in case $r$ is not dependant of $\theta$ we have this formula, else it would be misleading for people thinking "polar coordinates". This is the reason I deleted my previous comment and formulated better. $\endgroup$
    – zwim
    Jul 2, 2020 at 19:52

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