Compute (directional) derivative $u \in C^2(\mathbb{R}^2)$ and $v(r,\phi):=u(r\cdot \cos(\phi),r\cdot \sin(\phi))$. I have to show:
$u_{xx} + u_{yy} = v_{rr} + \frac{1}{r}v_r + \frac{1}{r^2}v_{\phi\phi}$
Where $x =r\cdot cos(\phi)$ and $y = r \cdot sin(\phi))$.
My problem is that I do not know how to compute for example $u_x$ or $u_v$.
 A: Using the chain rule,
$$ v_r = u_x x_r + u_y y_r = u_x \cos{\phi} + u_y \sin{\phi} \\
v_{\phi} = u_x x_{\phi} + u_y y_{\phi} = r(-u_x \sin{\phi} + u_y \cos{\phi})  $$
and then for the second derivatives,
\begin{align} 
v_{rr} &= u_{xr} \cos{\phi} + u_{yr} \sin{\phi} \\
&= u_{xx} \cos^2{\phi} + u_{xy}\cos{\phi}\sin{\phi} + u_{yx}\sin{\phi}\cos{\phi} + u_{yy} \sin^2{\phi} \\
&=  u_{xx} \cos^2{\phi} + 2u_{xy}\cos{\phi}\sin{\phi} + u_{yy} \sin^2{\phi} \\
v_{\phi\phi} &= r(-u_{x\phi} \sin{\phi} -u_x \cos{\phi} + u_{y\phi} \cos{\phi} - u_y \sin{\phi}) \\
&= r^2(u_{xx} \sin^2{\phi} - u_{xy} \sin{\phi}\cos{\phi} - u_{yx} \cos{\phi} \sin{\phi} + u_{yy} \cos^2{\phi} )  -r(u_x \cos{\phi}  + u_y \sin{\phi}) \\
&= r^2(u_{xx} \sin^2{\phi} - 2u_{xy} \sin{\phi}\cos{\phi} + u_{yy} \cos^2{\phi} )  -r(u_x \cos{\phi}  + u_y \sin{\phi}),
\end{align}
so
$$ v_{rr}+\frac{1}{r^2}v_{\phi\phi} = u_{xx}(\cos^2{\phi}+\sin^2{\phi}) + u_{yy} (\sin^2{\phi}+\cos^2{\phi}) - \frac{1}{r}(u_x \cos{\phi}  + u_y \sin{\phi}) \\
= u_{xx} + u_{yy} - \frac{1}{r}u_r $$
