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

From Wolfram Mathworld, we see the following relations:

$$\begin{align} \hat{\boldsymbol\rho} &= \cos\phi\hat{\mathbf x}+\sin\phi\hat{\mathbf y} \\ \hat{\boldsymbol\phi}&=-\sin\phi\hat{\mathbf x}+\cos\phi\hat{\mathbf y} \\ \hat{\mathbf z}&=\hat{\mathbf z} \end{align}$$

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

$$\begin{align} \rho&=\sqrt{x^2+y^2} \\ x&=\rho\cos\phi \\ y&=\rho\sin\phi \\ \end{align}$$

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

$$\begin{align} \cos\phi&=\frac{x}{\sqrt{x^2+y^2}} \\ \sin\phi&=\frac{y}{\sqrt{x^2+y^2}} \\ \end{align}$$

And hence:

$$\begin{align} \hat{\boldsymbol\rho} &= \frac{x}{\sqrt{x^2+y^2}}\hat{\mathbf x}+\frac{y}{\sqrt{x^2+y^2}}\hat{\mathbf y}&= \frac{ x \hat{\mathbf x} + y \hat{\mathbf y}}{\sqrt{x^2+y^2}} \\ \hat{\boldsymbol\phi}&=-\frac{y}{\sqrt{x^2+y^2}}\hat{\mathbf x}+\frac{x}{\sqrt{x^2+y^2}}\hat{\mathbf y} &= \frac{- y \hat{\mathbf x} + x \hat{\mathbf y}}{\sqrt{x^2+y^2}}\\ \hat{\mathbf z}&=\hat{\mathbf z} \end{align}$$

As was to be demonstrated.