Where did they get this extra $\dot r$ from in this derivate? I'm looking at this math example and I don't understand where they got the $\dot r$ from?
$$ m \frac{d}{dt}(r^2\dot\theta)$$
$$ = m(2r\dot r\dot\theta + r^2 \ddot\theta).$$
I understand they used the product rule to get this derivative, I just don't understand where the extra $\dot r$ came from.
 A: $$
m\frac{d}{dt}(r^2\dot \theta) = m\left(\big(\frac{d}{dt}(r^2)\big)\,\dot\theta\,+\,r^2\frac{d}{dt}\big(\dot\theta\big)\right) \\
= m\left(2r\dot r\dot\theta\,+\,r^2\ddot\theta\right)$$
The extra $\dot r$ comes from the fact that $r$ is a function of time, i.e. $r=r(t)$, thus when differentiating $r^2$, we have to apply the product rule for differentiation:
$$
\frac{d}{dt}(r^2(t))=2r(t)\frac{d}{dt}(r(t)) \\
=2r\dot r$$
A: $ r $ is a function of $ t $, we mustn't forget that when we want to calculate $ \frac{\mathrm{d}r^{2}}{\mathrm{d}t}\left(t\right)$.
First of all, we know that if $ g : E \rightarrow F $, and $ f : F\rightarrow G$ are differentiable functions, then $ f\circ g: E \rightarrow G$ is differentiable, and $ \left(f\circ g\right)'=g'\times f'\circ g $.
In our case, $ f : x\mapsto x^{2} $, its derivative is known to be $ f':x\mapsto 2x $.
Differentiating $ r^{2}=f\circ r $ applying that rule gives the following : $$ \frac{\mathrm{d}\left(f\circ r\right)}{\mathrm{d}t}\left(t\right)=\color{red}{\frac{\mathrm{d}r}{\mathrm{d}t}\left(t\right)}\times f'\left(r\right)=\color{red}{\dot{r}}\times 2r=2r\dot{r} $$
