# Partial derivatives of the radius

We take partial derivatives of the radius with respect to Cartesian coordinates all the time, especially in physics. While we can very easily do so with the quotient / chain rule, I’ve always been annoyed by that process (e.g. dealing with fractional exponents / cancelling out the 2s). As such, below is a proof by picture which is hopefully more illuminating / less frustrating.

Suppose we have the vector \(\mathbf{r}\) living in \(\mathbb{R}^2\) decomposed into its components \(\mathbf{r} = x \mathbf{e}_x + y \mathbf{e}_y\), where \(\mathbf{e}_x = (1,0)\) and \(\mathbf{e}_y = (0,1)\). Let’s define \(r = \sqrt{x^2 + y^2} = \|\mathbf{r}\|\), the length of \(\mathbf{r}\).

To find \(\partial_x r\), instead of using calculus, we can draw the following picture.

We notice that as \(h \to 0\), we have \(\|\mathbf{r} + h \mathbf{e_x}\| \to r + h \partial_x(r)\). Taking this approximation, we have

\[\cos \theta = \frac{x}{r} = \partial_x(r)\]Similarly, we can conclude

\[\sin \theta = \frac{y}{r} = \partial_y(r)\]These results further inform us of the gradient of the radius.

\[\begin{align*} x\mathbf{e_x} + y \mathbf{e_y} &= r (\partial_x(r) \mathbf{e_x} + \partial_y(r) \mathbf{e_y}) \\ \therefore \quad \mathbf{r} &= r \nabla r \end{align*}\] Next maths post Previous maths postAll maths posts