If we consider the the relativistic Lorentz force law:

$$\frac{d}{dt} (m\gamma \vec{u})=e(\vec{E}+\vec{u} \times \vec{B})$$

How can we deduce:

$$\frac{d}{dt} (m\gamma c^2)=e \vec{E} \cdot \vec{u}$$

Clearly dotting with $\vec{u}$ will give us the RHS. Which leaves us:

$$\vec{u} \cdot \frac{d}{dt} (m\gamma \vec{u})=e \vec{u} \cdot \vec{E}$$

Could anyone help explain how to proceed and if this is the correct method?

If we consider the the relativistic Lorentz force law:

$$\frac{d}{dt} (m\gamma \vec{u})=e(\vec{E}+\vec{u} \times \vec{B})$$

How can we deduce:

$$\frac{d}{dt} (m\gamma c^2)=e \vec{E} \cdot \vec{u}$$

Clearly dotting with $\vec{u}$ will give us the RHS. Which leaves us:

$$\vec{u} \cdot \frac{d}{dt} (m\gamma \vec{u})=e \vec{u} \cdot \vec{E}$$

Could anyone help explain how to proceed and if this is the correct method?

EDIT: If it helps: with reference to these notes i'm working through: http://www.maths.ox.ac.uk/system/files/coursematerial/2012/2393/8/WoodhouseLectures.pdf Page 86, eq (178), the paragraph underneath states 'The first equation (which follows from the second)', this is what i'm trying to prove (a warning, the notes are riddled with errors..).