Given that $x =r\cos \theta$ and $y = r\sin\theta$, determine $\frac{\partial r}{\partial x}$ and $\frac{\partial \theta}{\partial x}$
Attempt:
I thought I'd calculate $\frac{\partial x}{\partial r}$, treat it as a fraction and say $\frac{\partial r}{\partial x} = 1/\frac{\partial x}{\partial r}$, however that doesn't work.
Then I thought I'd set $r = \frac{x}{\cos \theta}$ and differentiate about $x$, but this path leads to the same result above.
I'm not sure how $y = r\sin \theta$ comes into play here.
Official Answer:
$$ \frac{\partial r}{\partial x} = \cos \theta, \quad \frac{\partial \theta}{\partial x} = \frac{- \sin\theta }{r}$$
Observe that $r^2=x^2+y^2$, so $$2r\frac{\partial r}{\partial x}=2x \implies \frac{\partial r}{\partial x}=\frac{x}{r}=\cos\theta$$ Similarly you can use $\tan \theta=\frac{y}{x}$ for the other partial derivative.