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## Homework Statement

## \frac{d}{dt}\gamma(t)\vec{u(t)} ##

## Homework Equations

See above

## The Attempt at a Solution

This comes from trying to verify a claim in Chapter 12 of Griffiths Electrodynamics, 4th. edition (specifically Eq. 12.62 -> Eq. 12.63, if anyone has it on hand).

I would have expected that the usual product rule applies in this case, and that:

## \frac{d}{dt}\gamma(t)\vec{u(t)} = \frac{d\gamma(t)}{dt}\vec{u(t)} + \gamma(t)\frac{d\vec{u(t)}}{dt} ##

And that all vector operations on the result would proceed as normal (i.e. ## \frac{d}{dt}(\gamma(t)\vec{u(t)}) \cdot \vec{u(t)} = \frac{d\gamma(t)}{dt}\vec{u(t)} \cdot \vec{u(t)} + \gamma(t)\frac{d\vec{u(t)}}{dt} \cdot \vec{u(t)} ##).

But the text of the book appears to imply that:

## \frac{d}{dt}(\gamma(t)\vec{u(t)}) \cdot \vec{u(t)} = \frac{d\gamma(t)}{dt}\frac{d\vec{u(t)}}{dt} \cdot \vec{u(t)} ##

What am I doing wrong here? Does the product rule not apply when one of the quantities to be differentiated is a vector and the other a scalar? That just doesn't sound right.

Edit: The exact chain of equations from the textbook is:

## \frac{d\vec{p}}{dt} = \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = \frac{m\vec{u}}{(1-\frac{u^2}{c^2})^{\frac{3}{2}}} \cdot \frac{d\vec{u}}{dt} = \frac{d}{dt}(\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}) ##