# Derivative of kinetic energy [closed]

I read that the derivative of kinetic energy=$F\cdot v$. I tried to differentiate (1/2) mv^2 with respect to time but each time I am getting $m*v$ and not $m*a*v$ which solves to $F*v$. My efforts are as follows:

Required URL from where I found it: http://www.feynmanlectures.caltech.edu/I_13.html

Here is the procedure:

$KE = 0.5mv^2$

$\frac{d}{dt}KE = 0.5m\frac{d}{dt}v^2$

So the question becomes,how do we find the derivative of $v^2$ with respect to time?

One can easily see that $\frac{d}{dt} = \frac{dv}{dt}\frac{d}{dv}$ (Notice how the $dv$ cancels top and bottom)

Therefore,

$\frac{d}{dt}v^2 = \frac{dv}{dt}\frac{d}{dv}v^2 = \frac{dv}{dt}\times 2v$

Therefore, $\frac{d}{dt}KE = 0.5\times2mv\frac{dv}{dt} = vm\frac{dv}{dt} = vma = Fv$

• @mew-Is this called the chain rule? Commented Jul 3, 2015 at 15:34
• Yes. The chain rule is the part where d/dt = dv/dt times d/dv Commented Jul 3, 2015 at 15:36
• So is the rate of change of kinetic energy equal to power? But in definition of power is just the change of energy with respect to time. Commented Sep 18, 2020 at 23:37
• @AntoniosSarikas that's right, power is the rate of energy change, so if that's due to change in kinetic energy then power is Fv Commented Sep 19, 2020 at 3:06
• @Kenshin We have two forms of energy or potential. Can power refer to rate of change of potential energy? I am asking this because usually we have power connected to the transformation rate from one form to another or the rate at which work is done (which relates to the change of kinetic energy. So with what definition should we stick, $P=\frac{dE}{dt}$ or $P=\frac{dW}{dt}$ ? Commented Sep 19, 2020 at 15:05

Is this the correct way to find the derivative of kinetic energy?

$$K=\frac{1}{2}m v^2 \\$$ So: $$\frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt} v^2 + 2mv \frac{dv}{dt} \right)$$ If the mass does not change over the time, then $$\frac{dm}{dt}=0$$

And finally $$\frac{dK}{dt} = \frac{1}{2} \left(2mv \frac{dv}{dt} \right)$$

So simplifying: $$\frac{dK}{dt} = mv \frac{dv}{dt}=mav=F.v$$

• Thank for the correction docscience You're right!, K is a scalar Commented Jul 5, 2015 at 6:07

The time derivative of $v^2$ is $2v \frac{dv}{dt}$ not $2v$. You must use the chain rule.

• @mikestone-Can you please show me the steps? Commented Jun 29, 2015 at 13:22
• @soham Its basic calculus. Check chain rule on wikipedia. Commented Jun 29, 2015 at 13:25
• @soham If you check out any introductory calculus textbook, surely the chain rule figures prominently. It is an indispensable tool used for differentiating used in nearly all courses in calculus of any level. Commented Jun 29, 2015 at 13:53