Multivariable Calculus Help with Laplacian in Polar coordinates I am trying to see why 
$\big(\partial_{xx} + \partial_{yy}\big) u(r, \theta) = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta}$
I first use the chain rule to say that:
$\frac{\partial u}{\partial x} = u_r r_x + u_{\theta} \theta_x$
And then I calculate:
$r_x = \frac{x}{\sqrt{x^2 + y^2}} = \frac{rcos\theta}{r} = cos\theta$
$\theta_x = \frac{-y}{x^2+y^2} = \frac{-rsin\theta}{r^2} = \frac{-sin\theta}{r}$
Plugging in gives 
$\frac{\partial u}{\partial x} = u_r cos\theta - u_{\theta} \frac{sin\theta}{r}$
But I am unsure of how to take the next $x$ derivative and I am wondering if someone can help?

Let's start with
$$
\begin{cases}
x=r\cos(\theta)\\
y=r\sin(\theta).
\end{cases}
$$
We compute first $u_r:$
$$
u_r=u_xx_r+u_yy_r=\cos\theta u_x+\sin\theta u_y.
$$
$$
u_{rr}=\cos\theta u_{xr}+\sin\theta u_{yr}=\cos\theta (u_{xx}x_r+u_{xy}y_r)+sin\theta(u_{xy}x_r +u_{yy}y_r)=\\ 
=\cos^2\theta u_{xx}+2\cos\theta\sin\theta u_{xy}+\sin^2\theta u_{yy}.
$$
Similarly
$$
u_\theta=u_x x_\theta+u_yy_\theta=-r\sin\theta u_x+r\cos\theta u_y.
$$
So
$$
u_{\theta\theta}=-r(\cos\theta u_x+\sin\theta u_y)+r^2(\sin^2\theta u_{xx}-2\cos\theta\sin\theta u_{xy}+\cos^2\theta u_{yy}).
$$
Dividing both sides by $r^2$, adding $u_{\theta\theta}$ and $u_{rr}$, and rearraging term we obtain:
$$
\Delta u=u_{rr}+\frac 1{r^2} u_{\theta\theta}+\frac 1r u_r
$$
A bit of calculation omitted at the end, feel free to ask if something is unclear or wrong.
