So I came across this derivation in the book Classical Mechanics by Herbert Goldstein. I don't follow from the second step onwards. I understand that there's a dot product, but how do you compute it? More importantly, why is there a 2 in the denominator outside the integral?
1 Answer
$v^2$ means $\mathbf{v}\cdot\mathbf{v}$, and the derivative of a dot product obeys the product rule. This should be enough of a hint for you to understand the factor of $1/2$.
As for how to compute the dot product, that is irrelevant in understanding why this formula is true, but can be relevant in applying this formula to particular situations.
If you are studying this level of physics, at some point you learned that the dot product of two vectors
$$\mathbf{a}=a_x\hat{\mathbf{x}}+a_y\hat{\mathbf{y}}+a_z\hat{\mathbf{z}}$$
and
$$\mathbf{b}=b_x\hat{\mathbf{x}}+b_y\hat{\mathbf{y}}+b_z\hat{\mathbf{z}}$$
can be computed as
$$\mathbf{a}\cdot\mathbf{b}=a_xb_x+a_yb_y+a_zb_z.$$