# Derivative of velocity squared

1. Mar 27, 2009

### sin2beta

1. The problem statement, all variables and given/known data
This really isn't a specific homework problem. It is just something that I've never actually known the reason for. An example is in the derivation of:

$$\frac{dT}{dt} = F\dot v [\tex] In order to arrive at it, I replace T with [tex]1/2mv^2[\tex] and assume m is constant and then I have to do the product rule on v^2 and I'm never sure why I can't use the power rule. Any explanation is appreciated. On a side note, this makes sense to me when I substitute momentum in for one of the velocities and then do the product rule. I just don't understand why I can't leave it as velocity and use the power rule. Thanks! 2. Mar 27, 2009 ### CompuChip Well, v = v(t) is a function of t, and you are differentiating with respect to t. So when you differentiate v(t)^2 you use the chain rule with u = v(t) and get [tex]\frac{d(v(t))^2}{dt} = \frac{du}{dv} \frac{dv(t)}{dt} = (2v) (v'(t))$$
where du/dv = d(v^2)/dv = 2 v, just like differentiating
$$(x^2 + 6)^2$$
with respect to x first gives you 2(x^2 + 6) by the power rule and then another factor of 2x because of the chain rule.

So, using v'(t) = a(t), and F = m v(t)
$$\frac{dT}{dt} = \frac12 m (2 v a) = m v a = F v$$
[If you want to treat v(t) as a vector, you need to be a little more careful as you will get some dot products]