$\frac{d(v^2)}{dx} = \frac{d((dx/dt)^2)}{dx}$ Derivative of Velocity Squared with respect to Position

Question

$$\frac{d(v^2)}{dx} = d\frac{((dx/dt)^2)}{dx}$$

Physically it makes sense - how does velocity squared change with respect to its position.

What would the analytical solution be?

$$\frac{d((dx/dt)^2)}{dx} = \frac{dx}{dt}\,d\frac{(dx/dt)}{dx} = ?$$

Answer 1

Using the chain rule gives

$$\begin{align} \frac{dv^2}{dx}&=\frac{dv^2}{dt}\frac{dt}{dx}\\\\ &=\frac{\color{blue}{\frac{dv^2}{dt}}}{\color{red}{\frac{dx}{dt}}}\\\\ &=\frac{\color{blue}{2v\frac{dv}{dt}}}{\color{red}{v}}\\\\ &=2a \end{align}$$

where $a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$ is the acceleration.