Help with derivation step in Kepler orbit, using chain rule

Question

Picture of equations

Hello, I am trying to work out the derivation of the Kepler orbit myself with help of the following wiki article, link:

In the picture above, I'm unsure how differentiating equation 5 with respect to $t$ produces equation 6. I tried using product rule: $$\frac{d}{dt}\left(\frac{dr}{dt}\right)=\frac{d}{dt}\left(\frac{dr}{dθ}\cdot\frac{dθ}{dt}\right)$$ $$\frac{d^2r}{dt^2}=\left(\frac{dr}{dθ}\right)'\frac{dθ}{dt}+\frac{dr}{dθ}\left(\frac{dθ}{dt}\right)'$$ $$\frac{d^2r}{dt^2}=\frac{d^2r}{dθ^2}\left(\frac{dθ}{dt}\right)^2+\frac{dr}{dθ}\frac{dθ}{dt^2}$$

Precisely, I don't understand how $\displaystyle{\left(\frac{dr}{dθ}\right)'\frac{dθ}{dt}}$ becomes $\displaystyle{\frac{d^2r}{dθ^2}\left(\frac{dθ}{dt}\right)^2}$.

Answer 1

By the Chain Rule $$\frac{d}{dt}\left(\frac{dr}{d\theta}\right)=\frac{d\theta}{dt}\frac{d}{d\theta}\left(\frac{dr}{d\theta}\right).$$

Answer 2

In general, if $y=f(x)$, then $$\dot y = {dy\over dt}={dy\over dx}\cdot{dx\over dt}={dy\over dx}\dot x.$$ Here, let $x$ be $\theta$ and $y$ be $\frac{dr}{d\theta}$.