I'm having trouble understanding this derivation In my notes I came across an equation:
$$\ddot\theta = \frac{\mathrm d\dot\theta}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt}\left(\frac{v\sin(\theta)}{r}\right) = \frac{\dot v \sin (\theta) }{r} + \frac{v\dot \theta \cos(\theta)}{r} - \frac{v\sin(\theta)}{r^2}\dot r$$
The last term doesn't make sense to me. The two terms before that clearly come from the chain rule, but how was the third derived?
$v$ here is velocity, and $r$ is radius.
If this isn't enough information, even knowing that would be very helpful.
Thanks in advance for any help!
 A: For a fraction whose numerator and denominator are functions $x$ and $y$ of $t$, $\large \frac{x}{y}$, using the product rule (also called the quotient rule?) the derivative is given by:-
$\large \frac{d}{dt}(\frac{x}{y})=\frac{y\dot{x}-x\dot{y}}{y^2}$
If we let $\large x=vsin(\theta)$ and $\large y=r$, then
$\large \dot{x}=\dot{v}sin(\theta)+v\dot{\theta}cos(\theta)$, 
where for the second term, we use the chain rule
$\large \frac{d}{dt}sin(\theta)=\frac{d}{d\theta}(sin(\theta))\frac{d\theta}{dt}=\dot{\theta}cos(\theta)$
and
$\large \dot{y}= \frac{d}{dt}r=\dot{r}$
thus
$\Large \frac{d}{dt}\frac{vsin(\theta)}{r}=\frac{r(\dot{v}sin(\theta)+v\dot{\theta}cos(\theta))-vsin(\theta)\dot{r}}{r^2}=\frac{\dot{v}sin(\theta)}{r}+\frac{v\dot{\theta}cos(\theta)}{r}-\frac{vsin(\theta)}{r^2}\dot{r}$
A: Consider it as a ternary product:
$$\left(v\cdot\sin(\theta)\cdot\frac1r\right)^\bullet =  v^\bullet\sin(\theta)\frac1r+v\cos(\theta)\theta^\bullet \frac1r+v\sin\theta\left(-\frac{r^\bullet}{r^2}\right)$$
