I need to show that next equation stands:
$\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = \frac{1}{r}\frac{\partial}{\partial r} (r\frac{\partial u}{\partial r} ) +\frac{1}{r^2}(\frac{\partial^2 u}{\partial \theta^2} )$ where $u=f(x,y)$ , $x = r\cos\theta$ , $y = r\sin\theta$ , $r = \sqrt{x^2+y^2}$ , $\theta = \arctan(\frac{y}{x})$
I got lost where I came around $du=dx(\frac{\partial u}{\partial r}\frac{x}{r} - \frac{\partial u}{\partial \theta}\frac{y}{r^2})+dy(\frac{\partial u}{\partial r}\frac{y}{r} + \frac{\partial u}{\partial \theta}\frac{x}{r^2})\\\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{x}{r} - \frac{\partial u}{\partial \theta}\frac{y}{r^2},\ \ \ \ \ \ \ \frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\frac{y}{r} + \frac{\partial u}{\partial \theta}\frac{x}{r^2}$
I just can't figure out, what is the next step and how do I go on with this. Could you, please, help me?