# A derivative is puzzling me | Liénard-Wiechert E field derivation

I'm following the derivation for the Field of a moving point charge, from Griffiths' Introduction to Electrodynamics. Particularly, on page 458 the author states that

I find this familiar except for the $$\frac{1}{2}$$ that vanished from the first line. I know the 2nd line comes from a triple product but I can't see why there's a $$2$$ factor and where does it come from.

• Aren't these identities mentioned on the inside covers and the first chapter of that book? Nov 20, 2020 at 3:59

The full product rule for the gradient of a dot product is $$\nabla(A\cdot B) = (A\cdot\nabla)B + (B\cdot\nabla)A + A \times(\nabla\times B) + B \times(\nabla\times A)$$ where I've elided the vector symbols. This means that in $$\nabla(r'\cdot r')$$, there will be two terms each of $$(r'\cdot\nabla)r'$$ and $$r'\times(\nabla\times r')$$, hence the factor of $$2$$.

This is a common identity in vector calculus for the gradient of a dot product. For two vectors $$\mathbf{A}$$ and $$\mathbf{B}$$, we have: $$\nabla(\mathbf{A}\cdot\mathbf{B})=(\mathbf{A}\cdot\nabla)\mathbf{B}+(\mathbf{B}\cdot\nabla)\mathbf{A}+\mathbf{A}\times(\nabla\times\mathbf{B})+\mathbf{B}\times(\nabla\times\mathbf{A})$$

Note what happens in the case where $$\mathbf{A}=\mathbf{B}$$: \begin{align} \nabla(\mathbf{A}\cdot\mathbf{A})&=(\mathbf{A}\cdot\nabla)\mathbf{A}+(\mathbf{A}\cdot\nabla)\mathbf{A}+\mathbf{A}\times(\nabla\times\mathbf{A})+\mathbf{A}\times(\nabla\times\mathbf{A})\\ &= 2\left[(\mathbf{A}\cdot\nabla)\mathbf{A}+\mathbf{A}\times(\nabla\times\mathbf{A})\right] \end{align}