# Partial Derivative and Chain Rule

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!

• Do you know the general formula: $$\frac{\mathrm{d} f(\alpha(t), \beta(t))}{\mathrm{d} t}(t_0) = \frac{\partial f}{\partial x}(\alpha(t_0)) \cdot \frac{\mathrm{d} \alpha}{\mathrm{d} t}(t_0) + \frac{\partial f}{\partial y}(\beta(t_0)) \cdot \frac{\mathrm{d} \beta}{\mathrm{d} t}(t_0) \ ?$$ – Adayah Aug 5 '17 at 16:45
• No I do not. This formula did not appear in my Real Analysis lecture. Can you derive this formula for me or provide a link? – elfeck Aug 5 '17 at 16:54

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 $$f_x(r\cos\phi,r\sin\phi)\cos\phi+f_y(r\cos\phi,r\sin\phi)\sin\phi\ .$$
• (+1) Very nice explanation, although I'm not sure if the OP knows what the (total) derivative $df$ of a function $f : \mathbb{R}^2 \to \mathbb{R}^2$ is. – Adayah Aug 5 '17 at 19:04