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

Question

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.

Answer 1

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}$ in terms 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}(-r\sin(\theta)) + \frac {\partial u}{\partial y}(r\cos(\theta))$

Solve the above system for $ \frac {\partial u}{\partial x}$ and $ \frac {\partial u}{\partial y}$, and you will get the result.