Question about a harmonic oscillator integral

[Heath Watts]

Hi,
I'm trying to learn quantum physics (chemistry) on my own so that my work with Gaussian and Q-Chem for electronic structural modeling is less of a black box for me. I've reached the harmonic oscillator point in McQuarrie's Quantum Chemistry book and I'm having trouble justifying a step in his math. It's the integral of force with respect to x.

Integrate[m*(d2x/dt2), dx]
This says integrate the second derivative of time with respect to t for the integration variable x.

Changing the variable of integration to time gives:

Integrate[m*(d2x/dt2)*(dt/dt), dx]
or
Integrate[m*(d2x/dt2)*(dx/dt), dt]

Then something occurs here:
Integrate[m*(d(dx/dt)/dt)*(dx/dt), dt]

Integrate[(m/2)*d((dx/dt)*(dx/dt))/dt, dt]

Integrate[(m/2)*d((dx/dt)^2)/dt, dt]

What calculus rule have I forgotten that says that
(d2x/dt2)*(dx/dt)=(1/2)*d((dx/dt)^2)/dt

I can't seem to find it in any of my old textbooks or online. I hope that my notation is clear. I appreciate your help. If you can direct me to a website that explains this rule, I'd appreciate it.
Thanks,
Heath

 

[Nick89]

If you work backwards it's easy to see: it's just the chain rule!

$$\frac{d}{dt} \left(\frac{1}{2} \left( \frac{dx}{dt}\right)^2 \right) = 2 \times \left( \frac{1}{2} \frac{dx}{dt} \right) \times \frac{d}{dt} \frac{dx}{dt} = \frac{dx}{dt} \frac{d^2x}{dt^2}$$

Substituting $\frac{dx}{dt} = f(t)$ maybe helps:
$$\frac{d}{dt} \left( \frac{1}{2} \left( f(t) \right)^2 \right) = 2 \times \left(\frac{1}{2} f(t) \right) \times \frac{df}{dt} = f(t) \frac{df}{dt}$$

Or even easier in 'words':
The derivative of $f^2$ is $2f \, f'$. Of course, we don't want the two, so we use 1/2 in front of the $f^2$ term (which doesn't change the differentiation process since it's a constant).It's a pretty common differentiation 'trick' .

 

[Heath Watts]

Thanks very much Nick. The chain rule! How embarrassing.