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

From Wolfram Mathworld, we see the following relations:

$\begin{array}{cc}\hat{\mathit{\rho }}& =\cos \varphi \widehat{𝐱}+\sin \varphi \widehat{𝐲}\\ \hat{\mathit{\varphi }}& =-\sin \varphi \widehat{𝐱}+\cos \varphi \widehat{𝐲}\\ \widehat{𝐳}& =\widehat{𝐳}\end{array}$

Additionally, it has already been shown on the same page that:

$\begin{array}{cc}\rho & =\sqrt{x^2+y^2}\\ x& =\rho \cos \varphi \\ y& =\rho \sin \varphi \\ \end{array}$

Now, we simply need to combine these relations together to obtain:

$\begin{array}{cc}\cos \varphi & =\frac{x}{\sqrt{x^2+y^2}}\\ \sin \varphi & =\frac{y}{\sqrt{x^2+y^2}}\\ \end{array}$

And hence:

$\begin{array}{ccc}\hat{\mathit{\rho }}& =\frac{x}{\sqrt{x^2+y^2}}\widehat{𝐱}+\frac{y}{\sqrt{x^2+y^2}}\widehat{𝐲}& =\frac{x\widehat{𝐱}+y\widehat{𝐲}}{\sqrt{x^2+y^2}}\\ \hat{\mathit{\varphi }}& =-\frac{y}{\sqrt{x^2+y^2}}\widehat{𝐱}+\frac{x}{\sqrt{x^2+y^2}}\widehat{𝐲}& =\frac{-y\widehat{𝐱}+x\widehat{𝐲}}{\sqrt{x^2+y^2}}\\ \widehat{𝐳}& =\widehat{𝐳}\end{array}$

As was to be demonstrated.