$P: (0,\infty)\times \mathbb R \to \mathbb R^2, \quad P(r,\phi):=(r\cos\phi, r\sin\phi)$
$f: \mathbb R^2\setminus \{(0,0)\} \to \mathbb R$
$g(r,\phi):=f(P(r,\phi))$
I have to show that
$\partial_x f(P(r,\phi)) = \partial_r g(r,\phi)\cos(\phi)-\frac{1}{r}\sin(\phi)\partial_\phi g(r,\phi)$
I know that it is rather easy using the chain rule but I have a problem getting the substitution with polar coordinates here. I'd start like this:
$\partial_x f(P(r,\phi))=\partial_x [f(P(r,\phi))]\cdot \partial_x P(r,\phi)=\partial_x g(r,\phi)\cdot \partial_x P(r,\phi)$
My problem is, that I have no idea how to handle $\partial_x g(r,\phi)$ and $\partial_x P(r,\phi)$.
Can someone point me in the right direction?