# Is there an easier way to calculate the partial derivative?

If $$w=z\tan^{-1}\Big(\frac{x}{y}\Big),$$

find$$\frac{\partial^2 {w}}{\partial{x^2}}+\frac{\partial^2 {w}}{\partial{y^2}}+\frac{\partial^2 {w}}{\partial{z^2}}=?$$

I calculated this and the answer is zero. But It was long calculations ( find $$\frac{\partial {w}}{\partial{x}}$$ then calculate $$\frac{\partial^2 {w}}{\partial{x^2}}$$ and do this again for $$y$$ and $$z$$ ): \begin{align} \frac{\partial {w}}{\partial{x}}&=\frac{yz}{x^2+y^2}\\ \frac{\partial^2 {w}}{\partial{x^2}}&=\frac{2xyz}{(x^2+y^2)^2}\\ \frac{\partial {w}}{\partial{y}}&=\frac{-zx}{x^2+y^2}\\ \frac{\partial^2 {w}}{\partial{y^2}}&=\frac{-2xyz}{(x^2+y^2)^2}\\ \frac{\partial^2 {w}}{\partial{z^2}}&=0. \end{align}

I wonder if there is a shorter answer or another approach to calculate this? (Because the answer is zero I guess maybe there is different approach too.)

• it is obvious that $\frac{\partial^2 {w}}{\partial{z^2}}$ is zero, maybe the other two are easier to see after the first derivative, write out the details in the post – Arjang Feb 28 at 18:55
• Ok, I edited this – Soheil Feb 28 at 19:06
• Maybe recast the Laplacian in cylindrical coordinates? – Daniel Schepler Feb 28 at 19:23

The second derivative with respect to $$z$$ is clearly zero. Furthermore, $$\arctan(y/x) = \operatorname{Im}(\operatorname{Log}(x + \mathrm iy))$$ and we know that the imaginary part of an analytic function is harmonic (satisfies Laplace's equation). The result for $$\arctan(x/y)$$ follows by symmetry.
• How we know: $\arctan(y/x) = \operatorname{Im}(\operatorname{Log}(z))$? – Soheil Feb 28 at 19:23
• Maybe better write $\operatorname{Log}(x+iy)$ since the variable name $z$ is already in use. – David K Feb 28 at 19:24