notation - Derivative with respect to a vector is a gradient?

I've encountered in some books (and even completed an exercise from the Goldstein by using it), a strange notation that seems to work exactly like a gradient, I have tried to look for an explanation but found none yet and I think I can reduce the time spent looking for my answer by posting this question, even if you give me some literature to check I will be greatly thankful:

Why $$\frac{\partial}{\partial\mathbf{r}}=\nabla$$

As I understand it the partial derivative with respect to a vector is like aplying the gradient. I don't know why it seem so odd to me the notion of differentiating something with respect to a vector.

Answer

It's purely notation.

Given a real-valued function $f(\mathbf r) = f(x^1, \dots, x^n)$ of $n$ real variables, one defines the derivative with respect to $\mathbf r$ as follows: \begin{align} \frac{\partial f}{\partial \mathbf r}(\mathbf r) = \left(\frac{\partial f}{\partial x^1}(\mathbf r), \dots, \frac{\partial f}{\partial x^n}(\mathbf r)\right) \end{align} so, by definition, $\partial f/\partial \mathbf r$ is a vector of functions that precisely equals $\nabla f$. You may also run into the notation $\nabla_{\mathbf r}f$ which means precisely the same thing.

An advantage of the notations $\partial f/\partial\mathbf r$ and $\nabla_\mathbf r$ is that they make explicit the symbol used to label the argument of the function, and this can sometimes stave off confusion.

Addendum. You may also come across the following notation. Let $\mathbf f(\mathbf r) = \mathbf f(x^1, \dots, x^n)$ be a real, $m$-component vector-valued function of $n$ real variables, then define its derivative with respect to $\mathbf r$ as the following matrix, often referred to as the Jacobian matrix: \begin{align} \frac{\partial\mathbf f}{\partial \mathbf r}(\mathbf r) = \begin{pmatrix} \frac{\partial f^1}{\partial x^1}(\mathbf r) & \cdots & \frac{\partial f^1}{\partial x^n}(\mathbf r) \\ \vdots & \ddots & \vdots \\ \frac{\partial f^m}{\partial x^1}(\mathbf r) & \cdots & \frac{\partial f^m}{\partial x^n}(\mathbf r) \\ \end{pmatrix} \end{align} Consider, for example, the function $\mathbf g(\mathbf r) = \mathbf r$. In this case, you can convince yourself that its derivative $\partial/\partial\mathbf r$ is the identity matrix; \begin{align} \frac{\partial\mathbf g}{\partial \mathbf r}(\mathbf r) = \begin{pmatrix} 1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \\ \end{pmatrix} \end{align}