How can the differential of $\mathbf r\cdot\mathbf r$ be written as $2\mathbf r\cdot d\mathbf r$? $\mathbf r$ is a vector. How can the differential of $\mathbf r\cdot\mathbf r$ be written as $2\mathbf r\cdot d\mathbf r$?  That is,
$$d(\mathbf r\cdot\mathbf r)=2\mathbf r\cdot d\mathbf r$$
How? Here, $\mathbf r$ is a vector.
 A: If $\mathbf{r} = (x_1,\dots,x_n)$ then
$$d(\mathbf{r} \cdot \mathbf{r}) = d\left( \sum_{i = 1}^n x_i^2 \right) = \sum_{i = 1}^n d\left(x_i^2\right) = \sum_{i = 1}^n 2x_idx_i = 2\mathbf{r} \cdot d\mathbf{r}. $$
A: I see that you did not assume (without proof) that the product rule for single variables works the same for vectors.
That is careful thinking, which is good.
To find the differential of $\mathbf r \cdot \mathbf r,$ you can
write $\mathbf r$ termwise:
$$ \mathbf r = \mathbf i x + \mathbf j y + \mathbf k z,$$
then compute $\mathbf r \cdot \mathbf r$ taking the terms of the vectors one at a time,
then take the differential of the resulting expression.
On the other hand you can find the differential $\mathrm d\mathbf r$
term by term, compute the inner product $\mathbf r\cdot\mathrm d\mathbf r,$
and multiply by $2.$ You should get the same result as for 
$\mathrm d(\mathbf r \cdot \mathbf r),$ proving the two are equal.

Update: In the other answers you can actually see these steps carried out
rather than merely hinted at.
A: AlkaKadri and Trevor Gunn's answers are given in terms of coordinates, and David K's is given in terms of a particular basis. For a more abstract proof, we can consider the limit as h goes to zero of $\frac{\mathbf r \cdot \mathbf r - (\mathbf r+h \mathbf u) \cdot (\mathbf r+h \mathbf u)}{h}$ where h is a scalar and $\mathbf u$ is an arbitrary vector. We can proceed must like the product rule for scalar; the $\mathbf r \cdot \mathbf r$ terms cancel out and we're left with 2($\mathbf r \cdot \mathbf u$)+h($\mathbf u \cdot \mathbf u$). Since we're taking the limit as h goes to zero, we can ignore the second term. Interpreting $\mathbf u$ as d$\mathbf r$, we get 2$\mathbf r\cdot d\mathbf r$.
A: The first thing to realize is that you can differentiate vector functions. In single variable calculus, we are interested in functions that have $1$ "input" and $1$ output. I see that now you're likely in a multivariate calculus course, although you're still only interested in functions that have $1$ input, functions can now have $3$ outputs. Namely, you have $\vec{r}(t) = \left[ \begin{array}{c} x(t) \\ y(t) \\ z(t) \end{array} \right]$. To differentiate $\vec{r}$ with respect to $t$, we simply differentiate each component with respect to $t$
$$\frac{d\vec{r}}{dt} = \left[ \begin{array}{c} x'(t) \\ y'(t) \\ z'(t) \end{array} \right]$$
$$\Rightarrow d\vec{r} = \left[ \begin{array}{c} x'(t) dt \\ y'(t) dt \\ z'(t) dt \end{array} \right]$$
$$\Rightarrow d\vec{r} = \left[ \begin{array}{c} dx \\ dy \\ dz \end{array} \right]$$
This, is how we have the differential of a vector.
Recall that the product rule applies for the dot product between two vector functions. That is,
$$\frac{d}{dt} ( \vec{r}(t) \cdot \vec{s}(t)) =  \frac{d\vec{r}(t)}{dt} \cdot \vec{s}(t) + \vec{r}(t) \cdot \frac{d\vec{s}(t)}{dt} $$
In the case where we have $\vec{r}(t) \cdot \vec{r}(t)$,
$$\frac{d}{dt} ( \vec{r}(t) \cdot \vec{r}(t)) =  \frac{d\vec{r}(t)}{dt} \cdot \vec{r}(t) + \vec{r}(t) \cdot \frac{d\vec{r}(t)}{dt} = 2 \vec{r}(t) \cdot \frac{d\vec{r}(t)}{dt}$$
$$\Rightarrow d(\vec{r} \cdot \vec{r}) = 2\vec{r} d\vec{r} $$
