Acceleration as a function of displacement

[Woolyabyss]

In one my classes my lecturer showed us the following derivation of acceleration as a function of displacement

dv/dt = v(dv/dx) = d(.5v^2)/dx

I understand how to get from the first to the second part. But I'm not sure how he got from the second part to the third. Its almost like he integrated v dv on the right hand side?

Any help would be appreciated.

 

[ShayanJ]

$\frac{dv}{dt}=v\frac{dv}{dx}=\frac{1}{2}2v\frac{dv}{dx}=\frac{d}{dx}(\frac{1}{2} v^2)$

 

[Woolyabyss]

Oh so you're really just reversing the chain rule