Let $f : \mathbb{R}^2 \to \mathbb{R}$ be twice continuously differentiable and define $F : (0, \infty) \times \mathbb{R} \to \mathbb{R} $ by $$ F(r, \varphi) = f(r\cos \varphi, r \sin \varphi). $$
Now I want to calculate the partial derivative $ \partial_r F $. My solution says $$ \partial_r F = \Big\langle \nabla f(r \cos \varphi, r\sin \varphi)~,~(\cos \varphi, \sin \varphi)^{T} \Big\rangle $$ and I can't understand why that is.
What does it mean to differentiate $ f(r\cos \varphi, r \sin \varphi) $ with respect to $ r $? Is it a partial derivative? How does the gradient come into play?
To me it looks like an application of a chain rule where we differentiate the arguments of $ f $, after "differentiating" $ f $ with respect to $ r $, whatever that means.
Can someone explain what is going on here? Thanks!
You have $F=f\circ{\rm rect}$ where ${\rm rect}$ is the map $${\rm rect}:\quad (r,\phi)\mapsto(r\cos\phi,r\sin\phi)\ .$$ The chain rule says that $dF=df\circ d{\rm rect}$, which means that matrixwise $[dF]$ is equal to the matrix product $[df]\cdot [d{\rm rect}]$. This can be typographically represented in various ways, one of them being $$\nabla F(r,\phi)=\nabla f(r\cos\phi,r\sin\phi)\left[\matrix{\cos\phi&-r\sin/phi \cr \sin\phi &r\cos\phi\cr}\right]\ .$$ Now ${\partial F\over\partial r}$ is the first component of the vector $\nabla F$, hence is given by $$f_x(r\cos\phi,r\sin\phi)\cos\phi+f_y(r\cos\phi,r\sin\phi)\sin\phi\ .$$