Explanation behind Second Derivative of a Parametric Equation Formula

Note $$\frac{d}{dx}\left(\frac{dy}{dt}\right)=\frac{dt}{dx}\frac{d}{dt}\left(\frac{dy}{dt}\right)=\frac{dt}{dx}\frac{d^2y}{dt^2}$$ and $$\frac{d}{dx}\left(\frac{dx}{dt}\right)=\frac{dt}{dx}\frac{d^2x}{dt^2}$$

Use the quotient rule $$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy/dt}{dx/dt}\right)=\frac{\frac{d}{dx}(\frac{dy}{dt})\frac{dx}{dt}-\frac{d}{dx}(\frac{dx}{dt})\frac{dy}{dt}}{(\frac{dx}{dt})^2}=\frac{\frac{dt}{dx}\frac{d^2y}{dt^2}\frac{dx}{dt}-\frac{dt}{dx}\frac{d^2x}{dt^2}\frac{dy}{dt}}{(\frac{dx}{dt})^2}$$ $$=\frac{\frac{d^2y}{dt^2}-\frac{d^2x}{dt^2}\frac{dy}{dx}}{(\frac{dx}{dt})^2}$$ Now multiply top and bottom of the fraction by $\frac{dx}{dt}$: $$=\frac{\frac{dx}{dt}\frac{d^2y}{dt^2}-\frac{dy}{dt}\frac{d^2x}{dt^2}}{(\frac{dx}{dt})^3}$$ This is the correct formula for the second derivative of a parametric equation.

$f’(x)=\frac{dy}{dx}$ is a function of $x$, applying the chain rule gives:

$$\frac{df’}{dx} \frac{dx}{dt}=\frac{df’}{dt}=\frac{d}{dt} f’ $$

Now solve for $f’’(x)= \frac{df’}{dx} $ assuming $\frac{dx}{dt} \neq 0$.