Simple classical physics inquiry

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shanepitts
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Not sure how the extra velocity quantity appears after deriving both side of the velocity function to get acceleration. Please help.
 
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You're given v as a function of x, not as a function of t. Therefore you have to use the chain rule: $$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt}$$
 
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