Is there an easier way to prove this differentiation?

Question

Given that $$\begin{aligned} &x=r\cos(\theta),\\ &y=r\sin(\theta),\\ & \qquad \text{and}\\ &x^2+y^2=r^2 \end{aligned}$$ Show that $\partial_y(\theta)=\cos(\theta)/r.$

I tried the following:

Since $$\partial_\theta(y)=\partial_\theta(r)\sin(\theta)+r\cos(\theta)$$ then $$\partial_\theta(r)=\frac{\partial_\theta(y)-r\cos(\theta)}{\sin(\theta)}$$ Using $x^2+y^2=r^2:$ $$\partial_y\left( x^2+y^2\right)=\partial_\theta (r^2)\cdot\partial_y(\theta)\implies 2y=2r\,\partial_\theta (r)\cdot\partial_y(\theta)$$ Substituting and simplifying $$1=\frac{1}{\sin(\theta)}\cdot\frac{\partial_\theta(y)-r\cos(\theta)}{\sin(\theta)}\cdot\partial_y(\theta)\implies\partial_y(\theta)\cdot\frac{r\cos(\theta)}{\sin^2(\theta)}=\frac{1}{\sin^2(\theta)}-1$$ simplifying again $$\partial_y(\theta)=\frac{\cos(\theta)}{r}$$

Is there a quicker way to prove this result?

Answer 1

$$\theta=\arctan (\frac {y}{x}) +k\pi$$

$$\implies \;\frac {\partial \theta}{\partial y}=\frac {1}{x}\frac {1}{1+(\frac {y}{x})^2} $$ $$=\frac {x}{r^2} $$ $$=\frac {\cos (\theta)}{r}$$

Answer 2

$\begin{bmatrix} \frac {\partial x}{\partial r}&\frac {\partial y}{\partial r}\\ \frac {\partial x}{\partial \theta}&\frac {\partial y}{\partial \theta}\end{bmatrix} = \begin{bmatrix} \cos\theta&\sin\theta\\ -r\sin\theta&r\cos\theta\end{bmatrix}$

$\begin{bmatrix} \frac {\partial r}{\partial x}&\frac {\partial \theta}{\partial x}\\ \frac {\partial r}{\partial y}&\frac {\partial \theta}{\partial y}\end{bmatrix} =\begin{bmatrix} \frac {\partial x}{\partial r}&\frac {\partial y}{\partial r}\\ \frac {\partial x}{\partial \theta}&\frac {\partial y}{\partial \theta}\end{bmatrix}^{-1}= \begin{bmatrix} \cos\theta&\sin\theta\\ -r\sin\theta&r\cos\theta\end{bmatrix}^{-1} = \begin{bmatrix} \cos\theta&-\frac{\sin\theta}{r}\\ \sin\theta&\frac {\cos\theta}{r}\end{bmatrix}$