# How does this simplification of this derivative work?

So there is a simplification I don't understand it goes like this:

\begin{alignat}{2} &\text{Original equation:}\qquad 2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}&=0\\ &\text{Simplification:}\qquad\frac{1}{2r}\frac{d}{dt}\left(r^{2}\left(\frac{d\theta}{dt}\right)\right)&=0 \end{alignat}

The reason why I don't understand it: if I try to differentiate the simplification I won't get the original equation. When I differentiate it I get: $$\frac{1}{2r}\left(2r\left(\frac{d\theta}{dt}\right)+r^2\left(\frac{d^2\theta}{dt}\right)\right)$$ I'm just using the product rule. Could someone explain to me what I'm doing wrong. It would be very much appreciated.

Thanks in advance.

## 2 Answers

You applied the product rule incorrectly. You should have $$(uv)'=u'v+uv'$$ where $$u=r^2$$ and $$v=d\theta/dt$$.

Note that $$u'=2r r'\quad v'=\frac{d^2\theta}{dt^2}$$ where $$\displaystyle(\cdot)'=\frac{d(\cdot)}{dt}$$.

Notes.

\begin{align} \frac{1}{2r}\frac{d}{dt}\left(r^{2}\left(\frac{d\theta}{dt}\right)\right) &=\frac{1}{2r}(2r\frac{dr}{dt}\frac{d\theta}{dt}+r^2\frac{d^2\theta}{dt^2})\\ &=\frac{dr}{dt}\frac{d\theta}{dt}+\frac{r}{2}\frac{d^2\theta}{dt^2} \end{align}

• Thank you so much, I totally forgot that $(\cdot)^{\prime}=\frac{d(\cdot)}{d t}$, thanks – TheCreator Jan 19 at 16:50
• @TheCreator: you are welcome! – user9464 Jan 19 at 16:51

Careful, $$r$$ is also a function of $$t$$, so you should have: $$\frac{1}{2r}\frac{d}{dt}\left(r^{2}\left(\frac{d\theta}{dt}\right)\right) = \frac{1}{2r}\left(2r\frac{dr}{dt}\frac{d\theta}{dt} + r^{2}\frac{d^{2}\theta}{dt^2}\right) = \frac{dr}{dt}\frac{d\theta}{dt} + \frac{r}{2}\frac{d^{2}\theta}{dt^2}.$$ This gives us $$\frac{dr}{dt}\frac{d\theta}{dt} + \frac{r}{2}\frac{d^{2}\theta}{dt^2}=0,$$ which upon multiplying by $$2$$ gives us $$2\frac{dr}{dt} \frac{d\theta}{dt} + r\frac{d^{2}\theta}{dt^{2}} = 0.$$

• Thank you for your answer, I really appreciate it – TheCreator Jan 19 at 16:50