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Say we have a function

\begin{cases} f: \mathbb{R}^n \to \mathbb{R}\\ f(\mathbf{v}) = \mathbf{u}^\top \mathbf{v} \end{cases}

where $\mathbf{u} \in \mathbb{R}^n$.

Apparently taking the partial derivate of $f$ with respect to $\mathbf{v}$ yields $\mathbf{u}$:

$$\frac{\partial f}{\partial \mathbf{v}} = \mathbf{u}$$

Why is that? This makes no sense to me. As $f$ returns real numbers, the rate of change in $f$ should be a real number, I would have assumed. Why is the rate of change a vector? Vectors are not even part of co-domain of $f$.

Also, what subject do I need to look into for this? I just got confronted with that isolated claim that $\frac{\partial f}{\partial \mathbf{v}} = \mathbf{u}$, here.

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    $\begingroup$ The differential of a real function of several variables can't be a real number. $\endgroup$ Jun 22 '19 at 15:40
  • $\begingroup$ Have you heard of Jacobian matrix ? $\endgroup$
    – Jakobian
    Jun 22 '19 at 15:41
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$\frac {\partial f}{\partial \textbf v}$ is a shorthand for $\left(\frac {\partial f}{\partial v_1},...,\frac {\partial f}{\partial v_n}\right)$, in other words it is the gradient of $f$. In this case, if you expand the dot product notation in terms of the coefficients, you obtain $\frac {\partial f}{\partial v_i}=u_i$, so $\frac {\partial f}{\partial \textbf v} = \textbf u$.

In general, most rules for taking derivatives generalise well to taking derivatives with respect to vectors, as is done here, or even matrices. For a useful reference, I recommend the matrix cookbook, which has a list of identities. Proving a few might help you in your understanding.

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    $\begingroup$ Okay, so $f(v) = u \cdot v$, thus $\nabla f = \begin{bmatrix} \frac{\partial f}{\partial v_1}\\ \vdots\\ \frac{\partial f}{\partial v_n} \end{bmatrix} = \begin{bmatrix} \frac{\partial \sum_i^n u_i v_i}{\partial v_1}\\ \vdots\\ \frac{\partial \sum_i^n u_i v_i}{\partial v_n} \end{bmatrix} = \begin{bmatrix} u_1\\ \vdots\\ u_n \end{bmatrix} = u $ $\endgroup$ Jun 22 '19 at 15:59

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