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http://www.physics.usu.edu/Wheeler/QuantumMechanics/QMOrbitalAngularMomentum.pdf

Unfortunately, this is the first time I've come across a conversion to spherical coordinates and I'm pretty lost.

The set of equation at the bottom of page 2 is confusing me. Actually obtaining it via the chain rule is simple.

But how do I use it ? Specifically (and this is my major issue) how is $\frac{\partial r}{\partial x}=\frac{x}{r}$ ?

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  • $\begingroup$ What do you get if you differentiate $r=\sqrt{x^2+y^2+z^2}$ with respect to $x$? $\endgroup$ – Santiago Canez Apr 6 '14 at 15:51
  • $\begingroup$ I cannot imagine how I failed to notice this... Blame the quantum mechanics I've been doing lately. $\endgroup$ – Ari Ben Canaan Apr 6 '14 at 15:55
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$$\frac{\partial r}{\partial x}=\frac{\partial}{\partial x}\sqrt{x^2+y^2+z^2}=\frac{2x}{2\sqrt{x^2+y^2+z^2}}=\frac{x}{r}$$

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