From the chain rule of differentiation:

$$\frac{d(E^{2})}{dt} = \frac{d(E \cdot E)}{dt} = (E \cdot \frac{dE}{dt}) + (\frac{dE}{dt} \cdot E)$$ $$ \frac{d(E^{2})}{dt} = 2 \cdot E \cdot \frac{dE}{dt}$$ $$\frac{1}{2} \frac{d(E^{2})}{dt} = E \cdot \frac{dE}{dt}$$

Also known as the "BAC-CAB" or "back-of-the-cab" identity.

This may need to be further clarified as follows:

d/dt(D^2) = d/dt( D(t) * D(t) ) = D(t) * d/dt( D(t) ) + d/dt( D(t) ) * D(t) <--- (using chain rule)

This yields:

d/dt(D^2) = 2 * D(t) * d/dt( D(t) )