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 $$\partial_x r$$, we shall compare the length of $$\mathbf{r}$$ and $$\mathbf{r} + h \mathbf{e}_x$$ for some small $$h$$. 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 can draw the following picture.

Focus on the two similar right triangles in the above picture that share the same angle $$\theta$$, we have

$\cos \theta = \frac{x}{r} = \frac{h\partial_x(r)}{h} = \partial_x(r).$

Similarly, we can conclude

$\sin \theta = \frac{y}{r} = \partial_y(r).$

\begin{align*} \nabla r &= \partial_x(r) \mathbf{e}_x + \partial_y(r) \mathbf{e}_y \\ &= \frac{x}{r} \mathbf{e}_x + \frac{y}{r} \mathbf{e}_y \\ &= \frac{\mathbf{r}}{r} \end{align*}

Algebraic method

We can also easily calculate that $$\nabla(x^2 + y^2) = 2 (x,y)$$. So $$\nabla(r^2) = 2 \mathbf{r}$$. Now by chain rule we also have $$\nabla(r^2) = 2r \nabla r$$. Combining the two equations we get $$\nabla r = \mathbf{r}/r$$.