Coordinate systems/Derivation of formulas/Spherical from Cartesian unit vectors

From Wolfram Mathworld, we have the following relations for the unit vectors in a Spherical coordinate system:

${\begin{aligned}{\hat {\mathbf {r} }}&=\cos \phi \sin \theta {\hat {\mathbf {x} }}+\sin \phi \sin \theta {\hat {\mathbf {y} }}+\cos \theta {\hat {\mathbf {z} }}\\{\hat {\boldsymbol {\theta }}}&=\cos \phi \cos \theta {\hat {\mathbf {x} }}+\sin \phi \cos \theta {\hat {\mathbf {y} }}-\sin \theta {\hat {\mathbf {z} }}\\{\hat {\boldsymbol {\phi }}}&=-\sin \phi {\hat {\mathbf {x} }}+\cos \phi {\hat {\mathbf {y} }}\end{aligned}}$

Next, we need to change all the $\phi$ and $\theta$ to $x,y,z$. We use the following relations, also available on the same page:

${\begin{aligned}r&={\sqrt {x^{2}+y^{2}+z^{2}}}\\\theta &=\arccos({\frac {z}{r}})\\\phi &=\arctan({\frac {y}{x}})\end{aligned}}$

Using these, we can find the following quantities:

${\begin{aligned}\cos \theta &={\frac {z}{r}}={\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}\\\sin \theta &={\sqrt {1-\cos ^{2}\theta }}={\sqrt {1-{\frac {z^{2}}{x^{2}+y^{2}+z^{2}}}}}={\sqrt {\frac {x^{2}+y^{2}}{x^{2}+y^{2}+z^{2}}}}\\\cos \phi &={\frac {x}{\sqrt {x^{2}+y^{2}}}}\\\sin \phi &={\frac {y}{\sqrt {x^{2}+y^{2}}}}\end{aligned}}$

Lastly, we simply need to substitute these relations into our first set of equations to obtain our desired result:

${\begin{aligned}{\hat {\mathbf {r} }}&={\frac {x}{\sqrt {x^{2}+y^{2}}}}{\sqrt {\frac {x^{2}+y^{2}}{x^{2}+y^{2}+z^{2}}}}{\hat {\mathbf {x} }}+{\frac {y}{\sqrt {x^{2}+y^{2}}}}{\sqrt {\frac {x^{2}+y^{2}}{x^{2}+y^{2}+z^{2}}}}{\hat {\mathbf {y} }}+{\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}{\hat {\mathbf {z} }}={\frac {x{\hat {\mathbf {x} }}+y{\hat {\mathbf {y} }}+z{\hat {\mathbf {z} }}}{\sqrt {x^{2}+y^{2}+z^{2}}}}\\{\hat {\boldsymbol {\theta }}}&={\frac {x}{\sqrt {x^{2}+y^{2}}}}{\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}{\hat {\mathbf {x} }}+{\frac {y}{\sqrt {x^{2}+y^{2}}}}{\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}{\hat {\mathbf {y} }}-{\sqrt {\frac {x^{2}+y^{2}}{x^{2}+y^{2}+z^{2}}}}{\hat {\mathbf {z} }}={\frac {xz{\hat {\mathbf {x} }}+yz{\hat {\mathbf {y} }}-\left(x^{2}+y^{2}\right){\hat {\mathbf {z} }}}{{\sqrt {x^{2}+y^{2}}}{\sqrt {x^{2}+y^{2}+z^{2}}}}}\\{\hat {\boldsymbol {\phi }}}&=-{\frac {y}{\sqrt {x^{2}+y^{2}}}}{\hat {\mathbf {x} }}+{\frac {x}{\sqrt {x^{2}+y^{2}}}}{\hat {\mathbf {y} }}={\frac {-y{\hat {\mathbf {x} }}+x{\hat {\mathbf {y} }}}{\sqrt {x^{2}+y^{2}}}}\end{aligned}}$