why is this true?
$\frac{d\vec{r}}{dt} \cdot \frac{d^2\vec{r}}{dt^2} = \frac{1}{2} \frac{d}{dt}\left(\frac{d\vec{r}}{dt}\right)^2$
My problem states that $\vec{F}$ is a conservative field, ie: $F = \nabla \phi$ for some scalar potential $\phi$. $$\begin{align}\vec{F}&=m\vec{a}\\ &=m\frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} \end{align}$$
I now take the dot product of each side:
$$\vec{F} \cdot \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}= m\frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2}\cdot \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$$
Now, the textbook says that:
$$m\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \cdot \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} = \frac{m}{2} \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\right)^2$$
How did the textbook get this answer?: $\frac{m}{2} \frac{d}{dt}\left(\frac{d\vec{r}}{dt}\right)^2$ ?
All of this leads to the result, which I'm OK with:
$$\int \limits_{A}^{B} \vec{F} \cdot \mathrm{d}\vec{r} = \bigg[\frac{m}{2}v^2\bigg]^{B}_{A}$$
You can see it more easily by expanding the product. \begin{aligned} \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \cdot \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} & = \sum_{j=1}^n \frac{\mathrm{d}r_j}{\mathrm{d}t} \frac{\mathrm{d}^2r_j}{\mathrm{d}t^2} = \sum_{j=1}^n \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\mathrm{d}r_j}{\mathrm{d}t} \right)^2 \\ & = \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \sum_{j=1}^n \frac{\mathrm{d}r_j}{\mathrm{d}t} \frac{\mathrm{d}r_j}{\mathrm{d}t} = \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \cdot \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \right) = \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \right)^2 \end{aligned}