I Relativistic Work-Energy Theorem: Deriving Griffiths' Formula

[ArtVandolay]

TL;DR Summary

Stuck on derivation in Griffiths' Electrodynamics

In deriving the work-energy theorem, Griffiths does the following:

$\frac{d\mathbf{p}}{dt}\cdot\mathbf{u} = \frac{d}{dt}\bigg(\frac{m\mathbf{u}}{\sqrt{1-u^2/c^2}}\bigg)\cdot\mathbf{u}=\frac{m\mathbf{u}}{(1-u^2/c^2)^{3/2}}\cdot\frac{d\mathbf{u}}{dt}$

I may have forgotten something essential from vector calculus, but for the life of me, I can't figure out how he goes from the second term to the third. Using $\frac{d(u^2)}{dt}=\frac{d(\mathbf{u}\cdot\mathbf{u})}{dt}=2\mathbf{u}\cdot\frac{d\mathbf{u}}{dt}$, I'm getting:
$$\frac{d}{dt}\bigg(\frac{m\mathbf{u}}{\sqrt{1-u^2/c^2}}\bigg)\cdot\mathbf{u}=\bigg[\frac{m\frac{d\mathbf{u}}{dt}}{\sqrt{1-u^2/c^2}} - \frac{1}{2}\frac{m\mathbf{u}}{(1-u^2/c^2)^{3/2}}\bigg(-\frac{2\mathbf{u}}{c^2}\cdot\frac{d\mathbf{u}}{dt}\bigg)\bigg]\cdot\mathbf{u}$$
I can't see how to get this in the form shown above. What am I missing?

 

[PeterDonis]

What am I missing?

Bring the $\bf{u}$ inside the brackets so you have $\bf{u} \cdot d \bf{u} / dt$ in the first term and $\bf{u} \cdot \bf{u}$ in the second.

Then multiply the first term by $1 - u^2 / c^2$ top and bottom, and see what happens.

 

[ArtVandolay]

Then multiply the first term by $1 - u^2 / c^2$ top and bottom, and see what happens.

Ahh apparently I forgot how to do algebra. Thank you!