# How do you determine such derivatives can be simplified?

### Background

When working out the Laplacian in spherical coordinate system, either via chain rule or differentiating the basis vectors, one arrives at the following:

$\nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} + \frac{1}{r^2}\frac{\cos \theta}{\sin\theta}\frac{\partial}{\partial \theta} + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial \phi^2}$

which is further simplified to the following:

$\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta\frac{\partial}{\partial \theta}) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial \phi^2}$

### Question

I'm curious as to how one finds out the following simplification could be done:

$\frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r} = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})$

and

$\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} + \frac{1}{r^2}\frac{\cos \theta}{\sin\theta}\frac{\partial}{\partial \theta} = \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta\frac{\partial}{\partial \theta})$

Is there some technique or guideline to determine whether it is possible or not? I could easily prove it if I already know the final form, but if I have none, what can I use as a hint?

• I really like your question here. I hope to attack it from the perspective of metrics and musical isomorphisms when I have a bit of time. Perhaps someone will beat me to it, there is of course a plain calculus III answer to give here which is summarized briefly as because you can and/or the RHS of the expressions are total derivatives which are very desirable from the perspective of physics where integration of these set to zero immediately reveals conserved quantities.. nice question, I shall return. Commented May 28, 2014 at 14:49