Understating the acceleration equation in terms of velocity Sorry if this is considered a ‘basic’ question, but I’ve been stuck on this for a while now and can’t find much help. I know $\frac{d}{dx} \left(\frac{1}{2} v^2\right) =$ acceleration. But I do not understand what is wrong with my working out.
$$\left(\frac{d}{dt}\right)\left(\frac{dx}{dt}\right) = \left(\frac{d}{dx}\right)\left(\frac{dx}{dt}\right)\left(\frac{dx}{dt}\right) = \frac{d}{dx} (v^2)$$ I don’t know why I am missing the half that is in the original equation. I have seen the proof for it, but I do not exactly know what wrong assumption I am making. Thank you.
 A: Your working should go like this:
$a = \frac{dv}{dt}= \frac{dv}{dx} \cdot \frac {dx} {dt} = v\frac{dv} {dx}$
and at this point observe that $d(v^2) = 2vdv$, which allows you to rewrite the final expression as:
$ \frac{d(\frac 12 v^2)}{dx}$ as required.
Your error was in thinking $\frac {d}{dx} (\frac{dx} {dt})$ is a multiplication. It is not. It is differentiation of the velocity with respect to the displacement.
A: Just to add to Deepak's answer, the result he quotes in the first line actually holds for total derivatives as well. Perhaps this proof is useful to the OP
$$\frac{d v(x, t)}{dt}
= \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x} \frac{dx}{dt}
= \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x} v $$
$$\frac{d v(x, t)}{dx}
= \frac{\partial v}{\partial x} + \frac{\partial v}{\partial t} \frac{dt}{dx}
= \frac{\partial v}{\partial x} + \frac{\partial v}{\partial t} \frac{1}{v} $$
Multiplying lower equation by $v$ and subtracting the two yields
$$\frac{dv}{dt} = v \frac{dv}{dx}$$
