Q:

How do I prove this identity involving polar coordinates and $\nabla$?

Given that
$$\begin{aligned}
&x=r\cos(\theta),\\
&y=r\sin(\theta),\\
& \qquad \text{and}\\
&x^2+y^2=r^2
\end{aligned}$$
Use the chain rule to show that $\nabla=\mathbf{\hat{r}}\frac\partial{\partial r}+\mathbf{\hat{\theta}}\frac1r\frac\partial{\partial\theta}.$
I derived that $\,\mathbf{\hat{r}}=\cos(\theta)\mathbf{\hat{\text{i}}}+\sin(\theta)\mathbf{\hat{\text{j}}}\,$ and that  $\,\mathbf{\hat{\theta}}=-\sin(\theta)\mathbf{\hat{\text{i}}}+\cos(\theta)\mathbf{\hat{\text{j}}}.$ But I can't seem to gro from here. 

A:

In order to show  $$\nabla=\mathbf{\hat{r}}\frac\partial{\partial r}+\mathbf{\hat{\theta}}\frac1r\frac\partial{\partial\theta}.$$
we need to find $ \frac {\partial u}{\partial x}$ and $ \frac {\partial u}{\partial y}$interms of $r$ and $\theta.$
Note that $ \frac {\partial u}{\partial r}=     \frac {\partial u}{\partial x}   \frac {\partial x}{\partial r} +    \frac {\partial u}{\partial y}   \frac {\partial y}{\partial r} =  \frac {\partial u}{\partial x}cos(\theta) +   \frac {\partial u}{\partial y} sin(\theta)$
Similarly $ \frac {\partial u}{\partial \theta}=     \frac {\partial u}{\partial x}   \frac {\partial x}{\partial \theta} +    \frac {\partial u}{\partial y}   \frac {\partial y}{\partial \theta} =  \frac {\partial u}{\partial x}(-rsin(\theta)) +   \frac {\partial u}{\partial y}(rcos(\theta))$
Solve  the above system for $ \frac {\partial u}{\partial x}$ and $ \frac {\partial u}{\partial y}$, and you will get the result.

