confirm which one is correct? Let $f(z)=-(x^2+y^2)^{1/2}$ and $\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$. Help to confirm which one is correct for $\Delta f$; this or $(x^2+y^2)^{3/2}-\frac{2}{(x^2+y^2)^{1/2}}$?
thankyou.
EDIT: sorry, I meant above. Just one minus sign more.

This is an application of the chain rule.
$$\nabla^2\!f = \frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} . $$
If $f(x,y) = \sqrt{x^2+y^2}$ then the first order partial derivatives are:
$$\begin{array}{ccc}
\frac{\partial\! f}{\partial x} &=& \frac{x}{\sqrt{x^2+y^2}} \\
\frac{\partial\! f}{\partial y} &=& \frac{y}{\sqrt{x^2+y^2}}
\end{array}$$
Differentiating a second time, gives:
$$\begin{array}{ccc}
\frac{\partial^2 f}{\partial x^2} &=& \frac{y^2}{(x^2+y^2)^{3/2}} \\
\frac{\partial^2 f}{\partial y^2} &=& \frac{x^2}{(x^2+y^2)^{3/2}}
\end{array}$$
It follows that the Laplacian is given by
$$\begin{array}{ccc}
\nabla^2f &=& \frac{y^2}{(x^2+y^2)^{3/2}} + \frac{x^2}{(x^2+y^2)^{3/2}} \\
&=& \frac{x^2+y^2}{(x^2+y^2)^{3/2}} \\
&\equiv& \frac{1}{\sqrt{x^2+y^2}}
\end{array}$$
