If $f=F(x, y)$ and $x=re^\theta$ and $y=re^{-\theta}$, prove that $2x\frac{\partial{f}}{\partial{x}}=r\frac{\partial{f} } {\partial{r}} +\frac{\partial{f} } {\partial{\theta}} $
I'm not even making headway on this question. I need some clarifications on how the arbitrary function works. Am I to get rid of it or what? And hints on how this question can be solved will be greatly appreciated.
To solve the problem, you can view $x$ and $y$ as functions of $r, \theta$ and compute $\frac{\partial{f} } {\partial{r}}$ and $\frac{\partial{f} } {\partial{\theta}}$ in terms of $\frac{\partial{f}}{\partial{x}}$ and $\frac{\partial{f}}{\partial{y}}$ using the chain rule.
Then, substitute them in the RHS of your equation to get the result.
The Chain Rule states that
$\frac{df}{d\theta}=\frac{df}{dx}\frac{dx}{d\theta}+\frac{df}{dy}\frac{dy}{d\theta}$ and similarly $\frac{df}{dr}=\frac{df}{dx}\frac{dx}{dr}+\frac{df}{dy}\frac{dy}{dr}$.
We compute $\frac{df}{d\theta}=\frac{df}{dx}re^\theta-\frac{df}{dy}re^{-\theta}$
We compute $r\frac{df}{dr}=r\left[\frac{df}{dx}e^\theta+\frac{df}{dy}e^{-\theta}\right]=\frac{df}{dx}re^\theta+\frac{df}{dy}re^{-\theta}$.
Adding these together we get $2re^{\theta}\frac{df}{dx}=2x\frac{df}{dx}.$
