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So messing about with spherical coordinates, I wanted at some point to get the expressions for $\frac{\partial r}{\partial x}$ and $\frac{\partial x }{\partial r}$; using the convention:

$$ \left\{ \begin{array}{c} x = r \text{ cos} \theta \text{ sin} \phi \\ y = r \text{ sin} \theta \text{ sin} \phi \\ z = r \text{ cos} \phi. \\ \end{array} \right. $$

For $\frac{\partial x }{\partial r}$ I do:

$$ \frac{\partial{} }{\partial r} [r \text{ cos} \theta \text{ sin} \phi] = \text{ cos} \theta \text{ sin} \phi. $$

For $\frac{\partial r }{\partial x}$, I use:

$$ r^2 = x^2 + y^2 + z^2 $$ $$ \rightarrow \frac{\partial}{\partial x} [r^2] = 2 r \frac{\partial r }{\partial x} = 2 x $$ $$ \rightarrow \frac{\partial r }{\partial x} = \frac{x}{r} = \frac{r \text{ cos} \theta \text{ sin} \phi}{r} = \text{ cos} \theta \text{ sin} \phi. $$

How come I get the same for both?

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    Isn't $x = r \cos \theta \sin \phi$? – 2017-02-09
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    I measured $\theta$ from the x axis toward the y axis (like in the polar coordinates), and $\phi$ from the xy plane toward the position vector - comparing with the book I see they swap around $\phi$ and $\theta$ but they are just labels for the same thing as far as I can see – 2017-02-09
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    I urge you to use traditional notation, so that the limits are traditional as well. (The traditional limits on $\phi$ and $\theta$ are of course not the same. – 2017-02-09
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    I.e. I get $\text{ cos} \phi}$ because I measure $\phi$ from the "floor" instead of from the z-axis. But shouldn't it all add up anyways? – 2017-02-09
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    I'll edit the question to use traditional notation – 2017-02-09
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    Your fundamental problem is that you're using a non-standard definition of angles and then finding that you get a form of contradiction involving these angles for what should be a trivial pair of derivatives. Few are likely to help you when you use non-standard definitions in such a case, feeling that there is some trivial switching of variables that solves everything—that you should find on your own. – 2017-02-09
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    I changed it to use $\phi$ as you mention - I've checked to make sure he convention agrees with the one used in Wikipedia. But I get the same problem, the logic I used to get there is the same – 2017-02-09
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    Thanks for staying up on this one, am I doing something that is fundamentally wrong? To me they each seem like reasonable steps... – 2017-02-09

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