# Rearranging Derivatives

1. Sep 4, 2004

I've got this from a book. My question is simple. Does anyone know how to get to the expression (UNCLEAR) from Eq. (24.2)? I can verify it without any trouble, but the author does not explain how it was obtained in the first place.
THANK YOU VERY MUCH!!!

"Now let us consider the energy in a forced oscillator. The equation for the forced oscillator is
$$m\frac{d^2 x}{dt^2}+\gamma m \frac{dx}{dt}+m\omega _0 ^2 x = F(t).$$ (24.1)

In our problem, of course, $$F(t)$$ is a cosine function of $$t$$. Now let us analyze the situation: how much work is done by the outside force $$F$$? The work done by the force per second, i.e., the power, is the force times the velocity. $$\Big($$ We know that the differential work in a time $$t$$ is $$F dx$$, and the power is $$F\frac{dx}{dt}$$. $$\Big)$$ Thus

$$P=F\frac{dx}{dt}=m\left[\left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right)+\omega _0 ^2 x\left(\frac{dx}{dt}\right) \right]+\gamma m\left(\frac{dx}{dt}\right)^2 .$$ (24.2)

But the first two terms on the right can also be written as
$$\frac{d}{dt}\left[\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 +\frac{1}{2}m\omega _0 ^2 x^2 \right],$$ (UNCLEAR)
as is immediately verifyed by differentiating."

2. Sep 4, 2004

### Tide

Note that
$$\frac{d^2 x}{dt^2} = \frac {d}{dt} \frac {dx}{dt}$$
and, generally,
$$g \frac {dg}{dt} = \frac {1}{2} \frac {d g^2}{dt}$$
where g = g(t) is any function of t.

Last edited: Sep 4, 2004
3. Sep 5, 2004

Thank you for your reply, Tip. That $$g \frac {dg}{dt} = \frac {1}{2} \frac {d g^2}{dt}$$ pattern seems to fit!