Convert polar coordinate to Cartesian coordinate

Question

$$x=r\cos\theta,\,y=r\sin\theta,\;r^2=x^2+y^2,\,\theta=\arctan(y/x)$$ I was told that $\frac{\partial r}{\partial x}=\cos\theta,\,\frac{\partial r}{\partial y}=\sin\theta,\frac{\partial\theta}{\partial y}=\frac{\cos\theta}{r}$. I was trying to prove one of the partial derivatives $\frac{\partial r}{\partial x}$, but only to get tautology. Can some one point out where I made mistakes? $$1=\frac{\partial r}{\partial x}\cos\theta-r(\sin\theta)\frac{\partial\theta}{\partial x}\qquad\qquad(1)$$ $$\frac{\partial\theta}{\partial x}=\frac{1}{1+y^2/x^2}\frac{\partial (y/x)}{\partial x}=\cos^2\theta\frac{\frac{\partial y}{\partial x}x-y}{x^2}\qquad(2)$$ $$=\cos^2\theta\cdot\frac{\frac{\partial(r\sin\theta)}{\partial x}x-r\sin\theta}{x^2}=\frac{\cos^2\theta}{x^2}[(\frac{\partial r}{\partial x}\sin\theta+r(\cos\theta)\frac{\partial\theta}{\partial x})x-r\sin\theta]\qquad\qquad(3)$$ $$=\frac{\cos^2\theta}{x^2}[\frac{\partial r}{\partial x}x\sin\theta-r\sin\theta]+\frac{\partial\theta}{\partial x}\cos^2\theta\implies\frac{\partial\theta}{\partial x}=\frac{[\frac{\partial r}{\partial x}x\sin\theta-r\sin\theta]\cos^2\theta}{x^2\sin^2\theta}\qquad\qquad(4)$$ $$1=\frac{\partial r}{\partial x}\cos\theta-\frac{[\frac{\partial r}{\partial x}xr-r^2]\cos^2\theta}{x^2}=\frac{\partial r}{\partial x}\cos\theta-\frac{\partial r}{\partial x}\cos\theta+1\qquad\qquad(5)$$

Answer 1

I don't understand why you would begin with (1). As $r$ is positive by definition we have \begin{align} r=\sqrt{x^2+y^2} &\Rightarrow \frac{\partial r}{\partial x} = \frac{1}{2}(x^2+y^2)^{-\frac{1}{2}}\cdot2x & \text{(by the chain rule,)}\\ &\Rightarrow \frac{\partial r}{\partial x} = r^{-1}r\cos\theta & \text{(as $r^2=x^2+y^2$,)}\\ &\Rightarrow \frac{\partial r}{\partial x} = \cos\theta. & \end{align}