Using chain rule to find partial derivatives Let $ r = \sqrt {x^2+y^2}$ and $\theta= tan^{-1}(y/x)$ be the usual polar/rectangular relationships.  Furthermore, define $u(r(x,y),\theta(x,y)) = -sech^2(r)tanh(r)sin(\theta)$ and $v(r(x,y),\theta(x,y)) = sech^2(r)tanh(r)cos(\theta)$.
Show that $\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = sech^2(r)[sech^2(r)-2tanh^2(r)+\frac{tanh(r)}{r}] $
How would I go about doing this? I know I would have to use the chain rule, but how exactly would I go about finding $\frac{\partial v}{\partial x}$ and $\frac{\partial u}{\partial y}$?
Thanks for the help.
 A: Hint
$$x=r\cos\theta\quad,\quad y=r \sin\theta$$
$$\frac{\partial v}{\partial r}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial r}$$
$$-2{{\operatorname{sech}}^{2}}\,r\,{{\tanh }^{2}}r\cos \theta +{{\operatorname{sech}}^{4}}\,r\cos \theta=\cos \theta\frac{\partial v}{\partial x} +\sin \theta\frac{\partial v}{\partial y}\quad(1)$$
$$\frac{\partial v}{\partial \theta }=\frac{\partial v}{\partial x}\frac{\partial x}{\partial \theta }+\frac{\partial v}{\partial y}\frac{\partial y}{\partial \theta }$$
$$-{{\operatorname{sech}}^{2}}r\tanh \,r\,\sin \theta =-r\sin \theta \frac{\partial v}{\partial x}+r\cos \theta \frac{\partial v}{\partial y}\quad(2)$$
we have
$$\left\{ \begin{align}
  & \quad\,\,\,\cos \theta \frac{\partial v}{\partial x}\,\,+\,\,\,\,\,\sin \theta \frac{\partial v}{\partial y}=-2{{\operatorname{sech}}^{2}}r\,{{\tanh }^{2}}r\cos \theta +{{\operatorname{sech}}^{4}}r\cos \theta  \\ 
 & -r\sin \theta \frac{\partial v}{\partial x}\,+r\cos \theta \frac{\partial v}{\partial y}=-{{\operatorname{sech}}^{2}}r\tanh r\sin \theta  \\ 
\end{align} \right.$$
$$\frac{\partial v}{\partial x}=?$$
Similarly
$$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}$$
$$2{{\operatorname{sech}}^{2}}\,r\,{{\tanh }^{2}}r\sin \theta -{{\operatorname{sech}}^{4}}\,r\sin \theta=\cos \theta\frac{\partial u}{\partial x} +\sin \theta\frac{\partial u}{\partial y}\quad(3)$$
$$\frac{\partial u}{\partial \theta }=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta }+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta }$$
$${{\operatorname{sech}}^{2}}r\tanh \,r\,\cos \theta =-r\sin \theta \frac{\partial u}{\partial x}+r\cos \theta \frac{\partial u}{\partial y}\quad(4)$$
we have
$$\left\{ \begin{align}
  &\quad\,\,\,\cos \theta \frac{\partial u}{\partial x}+\,\,\,\,\sin \theta \frac{\partial u}{\partial y}=2{{\operatorname{sech}}^{2}}r{{\tanh }^{2}}r\sin \theta -{{\operatorname{sech}}^{4}}r\sin \theta  \\ 
 & -r\sin \theta \frac{\partial u}{\partial x}+r\cos \theta \frac{\partial u}{\partial y}={{\operatorname{sech}}^{2}}r\tanh r\cos \theta  \\ 
\end{align} \right.$$
$$\frac{\partial u}{\partial x}=?$$
A: Use the chain rule:
$$\dfrac{\partial v}{\partial x} = \dfrac{\partial v}{\partial r}\dfrac{\partial r}{\partial x} + \dfrac{\partial v}{\partial \theta}\dfrac{\partial \theta}{\partial x}$$
$$\dfrac{\partial u}{\partial y} = \dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial y} + \dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial y}$$
for example: $$\dfrac{\partial v}{\partial r}=\left( \text{sech}^4r-2\text{tanh}^2(r)\,\text{sech}^2r\right)\cos\theta$$
...etc
