$$\frac{d(v^2)}{dx} = d\frac{((dx/dt)^2)}{dx}$$
Physically it makes sense - how does velocity squared change with respect to its position.
What would the analytical solution be?
$$\frac{d((dx/dt)^2)}{dx} = \frac{dx}{dt}\,d\frac{(dx/dt)}{dx} = ?$$
$\frac{d(v^2)}{dx} = \frac{d((dx/dt)^2)}{dx}$ Derivative of Velocity Squared with respect to Position
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$\begingroup$
calculus
derivatives
physics
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0Actually, that relates work done with kinetic energy. For your interest, see another answer [**here**](http://physics.stackexchange.com/questions/310064/question-about-a-v-mathrm-dv-mathrm-dx/310072#310072) – 2017-02-08
1 Answers
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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.