Using the same method of conversion of laplacian from cartesian to spherical coordinates and changing $\psi$ to $u$, I am trying to finish the demostration of the spherical laplacian. However, I have a problem, as after adding up $\frac{\delta^{2}u}{\delta x^{2}}+\frac{\delta^{2}u}{\delta y^{2}}+\frac{\delta^{2}u}{\delta z^{2}}$ for the part of $u=u(\theta)$, which is the last I need to calculate.
I have: $$\frac{1}{r^2}\frac{\delta ^{2}u}{\delta \theta^{2}}-\frac{\delta u}{\delta \theta}\frac{2sin(\theta)cos(\theta)}{r^2}+\frac{\delta u}{\delta \theta}\frac{cos(\theta)}{r^2 sin(\theta)} $$ where I should have: $$\frac{1}{r^2}\frac{\delta ^{2}u}{\delta \theta^{2}}+\frac{cos(\theta)}{r^2 sin(\theta)}\frac{\delta u}{\delta \theta}$$
As you can see, in what I have done I had an extra term $$-\frac{\delta u}{\delta \theta}\frac{2sin(\theta)cos(\theta)}{r^2}$$
I don't know if I have to simplify more or I have derived badly. I request aid on this problem
I had derived $$\frac{\delta u}{\delta x}= \frac{\delta u}{\delta \theta} \frac{cos(\theta)cos(\phi)}{r} $$ $$\frac{\delta u}{\delta y}= \frac{\delta u}{\delta \theta} \frac{cos(\theta)sin(\phi)}{r} $$ $$\frac{\delta u}{\delta z}= \frac{\delta u}{\delta\theta} \frac{-sin(\theta)}{r} $$ And my problem maybe in calculating: $$\frac{\delta^2 u}{\delta x^2}= \frac{\delta}{\delta x}(\frac{\delta u}{\delta \theta} \frac{cos(\theta)cos(\phi)}{r} )$$ $$\frac{\delta^2 u}{\delta y^2}= \frac{\delta}{\delta y}(\frac{\delta u}{\delta \theta} \frac{cos(\theta)sin(\phi)}{r})$$ $$\frac{\delta^2 u}{\delta z^2}= \frac{\delta}{\delta z}(\frac{\delta u}{\delta\theta} \frac{-sin(\theta)}{r})$$
I had simplified as they do in conversion of laplacian from cartesian to spherical coordinates at the end of the question, so try to give the answer in cosines and sines, please. Thank you.
