How do I calculate numerically a tensor in polar coordinates? You can formulate the question also like this: What is the easiest way of calculating directed derivative of a function if its values are evaluated in a cartesian grid?
a) fit a (spline) surface, differentiate exactly
b) differentiate numerically in cartesian coordinates, project to angle
c) something else
The function at hand is a displacement vector ($u : \mathbb{R}^2 \to \mathbb{R}^2$ ) and I need the circumferential strain tensor component $$\varepsilon_{\theta\theta} = \frac{\partial u_\theta}{\partial \theta}$$. (Also the radial strain would is somewhat interesting.)
background:
In 2d strain tensor has four components: $\varepsilon_{i,j}$, where i and j is either taken from {x,y} (cartesian) or {$\theta$,r} (polar). The first index defines the direction of displacement and the second index the differentiation direction, i.e. $$\varepsilon_{i,j} = \frac{\partial u_i}{\partial j} $$
u here is the 2d displacement vector. In cartesian coordinates above is easy, in polar not so much.
 A: If we set 
$$
x=\rho\cos\theta\\
y=\rho\sin\theta
$$
then 
$$
\mathbf{e}_\rho  = \cos\theta\,\mathbf{e}_1+\sin\theta\,\mathbf{e}_2,\\
\mathbf{e}_\theta=-\sin\theta\,\mathbf{e}_1+\cos\theta\,\mathbf{e}_2.
$$
We have
$$
\nabla\otimes\mathbf{u}=\left(\mathbf{e}_\rho\frac{\partial}{\partial\rho}+\mathbf{e}_\theta\frac{1}{\rho}\frac{\partial}{\partial\theta}\right)\otimes(u_\rho\mathbf{e}_\rho+u_\theta\mathbf{e}_\theta)
$$
and then
$$
(\nabla\otimes\mathbf{u})_{\theta\theta}=\frac{1}{\rho}\left(\frac{\partial u_\theta}{\partial\theta}+u_\rho\right)
$$
where the second term arises from the derivative of $\mathbf{e}_\rho$ with respect to $\theta$, giving $\mathbf{e}_\theta$.
If one would to express such quantities with respect to the corresponding Cartesian ones, the following formulas could be useful:
$$
u_\rho  =\mathbf{u}\cdot\mathbf{e}_\rho  = u_1\cos\theta+u_2\sin\theta\\
u_\theta=\mathbf{u}\cdot\mathbf{e}_\theta=-u_1\sin\theta+u_2\cos\theta\\
\frac{\partial}{\partial\theta}=-\rho\sin\theta\frac{\partial}{\partial x}+\rho\cos\theta\frac{\partial}{\partial y}
$$
A: I'm not sure if this is what you are looking for, but,
so, if $(x,y)$ is given by $x=r\cos\varphi$ and $y=r\sin\varphi$, then
$$dx = dr\cdot \cos\varphi+r\cdot (-\sin\varphi)d\varphi $$
$$dy = dr\cdot \sin\varphi+r\cdot \cos\varphi\, d\varphi$$
$$r=\sqrt{x^2+y^2}\ \text{ and }\ \varphi = \arctan \frac yx \color{grey}{(+\pi)}$$
$$ dr=\frac x{\sqrt{x^2+y^2}}dx+\frac y{\sqrt{x^2+y^2}}dy$$
$$\frac{\partial f}{\partial r}(x,y)=\frac x{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x} + \frac y{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}$$
$$d\varphi = -\frac y{x^2}\cdot\frac{1}{(y/x)^2+1}dx + \frac1x\cdot\frac{1}{(y/x)^2+1}dy $$
$$...$$
