# Acceleration as a function of displacement

 HW Helper P: 3,353 I Think you have to use the result $$a=x''=\frac{d}{dx} (\frac{1}{2}v^2)$$. That result can be obtained as such - $$\frac{d}{dx} ((1/2) v^2) = \frac{d}{dv} ((1/2) v^2) \cdot \frac{dv}{dx}=v\frac{dv}{dx}$$ by the chain rule. Another application of the chain rule: $$v\frac{dv}{dx} = \frac{dx}{dt}\cdot\frac{dv}{dx} = \frac{dv}{dt}=a$$