Suppose you have a function $y = y(x(t))$. By taking the first derivative wrt t we would get:
$$ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} $$
I'm a bit stuck trying to determine the second derivative, but this is what I have right now:
$$ \frac{d^{2}y}{dt^{2}} = \frac{d^{2}y}{dx^{2}} \frac{dx}{dt} + \frac{dy}{dx}\frac{d^{2}x}{dt^{2}}$$
I know this isn't correct though but i'm not exactly sure what I am missing.
First use the product rule to obtain
$$\begin{align} \frac{d}{dt}\left(\frac{dy}{dx}\frac{dx}{dt}\right)&=\frac{dx}{dt}\frac{d}{dt}\left(\frac{dy}{dx}\right)+\frac{dy}{dx}\frac{d}{dt}\left(\frac{dx}{dt}\right)\\\\ &=\frac{dx}{dt}\frac{d}{dt}\left(\frac{dy}{dx}\right)+\frac{dy}{dx}\frac{d^2x}{dt^2}\tag1 \end{align}$$
Next, apply the chain rule to the second term on the right-hand side of $(1)$ to find
$$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{dy}{dx}\right)\frac{dx}{dt}\tag2$$
Substitution of $(2)$ in $(1)$ yields the coveted result
$$\begin{align} \frac{d}{dt}\left(\frac{dy}{dx}\frac{dx}{dt}\right)&=\frac{d^2y}{dx^2}\left(\frac{dx}{dt}\right)^2+\frac{dy}{dx}\frac{d^2x}{dt^2} \end{align}$$
And we are done!
