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In particular, I have $$\frac{\partial r}{\partial x} = \frac{\partial (x^2 + y^2 + z^2)^{1/2}}{\partial x} = \frac{x}{(x^2 + y^2 + z^2)^{1/2}}= \frac{x}{r}$$ does this mean $$\frac{\partial x}{\partial r} = \frac{r}{x}$$ ?
If it is, can we safely assume that this is so for all partial derivatives?

non-sensical
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1 Answers1

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Of course, an equality is an equality (so this statement is general). The equality $ \frac{\partial x(r,y,z)}{\partial r} = \frac{r}{x}$ is also true.

You can easy proof that in this case using $ x = \sqrt{r^2 - y^2 - z^2}$. You will get: $$ \frac{\partial x}{\partial r} = \frac{2r}{2{\sqrt{r^2 - y^2 - z^2}}} = \frac{r}{x} $$.

More general, for a intuitive proof, let as consider $f = f(x,y,z)$ (it can easy be generalized to a function with $n$ variables. I will also try to simplify the notation).

$$ \frac{\partial f}{\partial x} = \lim_{\Delta x \to 0 } \frac{ f(x + \Delta x , y, z) - f(x,y,z)}{\Delta x} =\lim_{\Delta x \to 0 } \frac{\Delta f}{\Delta x} $$

$$ \frac{\partial x}{\partial f} = \lim_{\Delta f \to 0 } \frac{ x(f + \Delta f , y, z) - x(f,y,z)}{\Delta f} =\lim_{\Delta f \to 0 } \frac{\Delta x}{\Delta f} $$

From the second equality we get: $$\frac{1}{\frac{\partial x}{\partial f}} = \lim_{\Delta f \to 0 } \frac{\Delta f}{\Delta x} =\lim_{\Delta x \to 0 } \frac{\Delta f}{\Delta x} = \frac{\partial f}{\partial x} $$

(In the last step we used that for $\Delta f \to 0 $ and $ y,z $ fixed also $\Delta x \to 0 $)

Icarus 369
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    I find this answer quite misleading. By $\partial x/\partial r$ one would normally mean $\partial x(r,\theta,\phi)/\partial r$, which is not $r/x$. – Hans Lundmark Aug 07 '16 at 05:32
  • Sorry if I made a mistake - I thogut about $\partial x/\partial r$ as $\partial x (r,y,z)/\partial r$, because I understood $r$ as a general function $r = r(x,y,z)$, and $x = x(r,y,z)$. – Icarus 369 Aug 07 '16 at 06:36
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    Well, what you have written is true, but it's quite unusual in applications to ever view $x$ as a function of $(r,y,z)$. By the way, in the last equation I think you forgot to take the reciprocal somewhere, since you didn't really mean to show that $\partial f/\partial x = \partial f/\partial x$, did you? – Hans Lundmark Aug 07 '16 at 08:13
  • Yes, I forgot to write down one expression in that line. Now I fixed it. Thanks. – Icarus 369 Aug 07 '16 at 08:38
  • Late comment 6 years later to @HansLundmark: I get $\partial x(r,\theta,\phi)/\partial r = \sin \theta \cos \phi$, then transform back to Cartesian coordinates as $\sin \theta = \sqrt{x^2 + y^2} / r$, $\cos \phi = x / \sqrt{x^2 + y^2}$ so $\partial x / \partial r = r / x$. – Antimon Sep 25 '22 at 16:12
  • @Antimon: That's $x/r$, not $r/x$. – Hans Lundmark Sep 25 '22 at 20:21
  • Oh yeah, you're right. Sorry for the brain fart. – Antimon Sep 25 '22 at 20:52