# Polar coordinate partial derivative

Let $x=r \cos \theta, y=r \sin \theta$. Which of the following expression is equal to $\left( \frac{\partial \theta}{\partial r} \right)_x$, the partial derivative of $\theta$ with respect to $r$, keeping $x$ constant.

A) $r^{-1}\tan \theta$

B) $-r^{-1} \tan \theta$

C) $r^{-1}\cot \theta$

D) $-r^{-1}\cot \theta$

E) $-r^{-1}\csc \theta$

This question seems wrong what is the relation between $\theta$ and $r$? This question for me similar to what is $\frac{\partial y}{\partial x}$

$x=r \cos \theta$

$x/r = \cos \theta$

$${\partial(x/r)\over \partial r} = {\partial \cos \theta \over \partial r}$$

$$-xr^{-2} = {\partial \cos \theta / \partial \theta \over \partial r/ \partial \theta}\tag{1}$$

$$xr^{-2} = {\sin \theta \over \partial r/ \partial \theta}$$

$$xr^{-2} = { \partial \theta \over \partial r }\sin\theta\tag{2}$$

$${x\over r \times r\sin \theta} = { \partial \theta \over \partial r }$$

$${x\over y}r^{-1} = { \partial \theta \over \partial r }$$

$$r^{-1}\cot \theta = { \partial \theta \over \partial r }$$

(1) - chain rule

(2) - inverse function rule.

When $x$ is held constant, $$r=\frac{x}{\cos\theta}\implies \frac{\partial r}{\partial \theta}=\frac{x\sin \theta}{\cos^2\theta}\implies \frac{\partial \theta}{\partial r}=\frac{\cos^2\theta}{x\ \sin \theta}.$$ Plugging in $x=r\cos \theta$ then gives $$\frac{\partial \theta}{\partial r}=\frac{\cos^2\theta}{r\cos\theta\ \sin \theta}=\frac{\cot\theta}{r},$$ option (c).