Minor issue with partial derivatives

Question

Please see the image below for my question concerning partial derivates of functions in polar co-ordinates:

Answer 1

While $\frac{\mathrm dx}{\mathrm dy} \cdot \frac{\mathrm dy}{\mathrm dx} = 1$ is true when $x$ is a single-variable function of $y$, this has to fail for partial derivatives of multi-variable functions.

From $x = r \sin \theta \cos \phi$ you have correctly computed that $\frac{\partial x}{\partial r} = \frac{x}{r}$, when $x$ is considered a function of $r,\theta,\phi$. However, the equation $x = r \sin \theta \cos \phi$ can't possibly tell you how $r$ changes with $x$ when $y$ and $z$ are fixed. It only tells you how $r$ changes with $x$ when $\theta$ and $\phi$ are fixed, but that is not what $\frac{\partial r}{\partial x}$ is, when $r$ is considered a function of $x,y,z$.

As a function of $x,y,z$ you have $r = \sqrt{x^2+y^2+z^2}$ which does indeed yield $\frac{\partial r}{\partial x} = \frac{x}{r}$.

Answer 2

rewrite $r = (x^2+y^2+z^2)^\frac{1}{2}$

now $\frac{\partial r}{\partial x}=\frac{\frac{1}{2}\cdot 2x}{(x^2+y^2+z^2)^\frac{1}{2}} = \frac{x}{r}$ as required

Your $\frac{\partial x}{\partial r}=\frac{x}{r}$ is correct but when you invert it you actually have $r$ as a function of $x,\theta$ and $\phi$. Try to keep strictly to one coordinate system at a time