How is this equality involving vectors and partial derivatives true?

Question

I feel embarrassed having to ask this, but I can't seem to make sense of the following vector operations. In my textbook, the following two equalities are stated:

$$\mathbf B \cdot \frac{\partial \mathbf B}{\partial t} = \frac{1}{2} \frac{\partial }{\partial t}(B^2)$$ and $$\mathbf E \cdot \frac{\partial \mathbf E}{\partial t} = \frac{1}{2} \frac{\partial }{\partial t}(E^2).$$

Presumably, B = B(x, y, z, t) and E = E(x, y, z, t), so I am confused as to how one would move the B and E vectors "inside" the partial derivative operation since they too are functions of time. And where does the factor of $\frac{1}{2}$ come from?

Answer 1

This is just a direct application of the chain rule for univariate functions: $\,(f \circ g)' = (f' \circ g) \cdot g'\,$.

With $f(u)=u^2, g(t)=B(t)$ it follows that $f'(u)=2\,u, g'(t) = B'(t)$ so the chain rule gives: $$\left(B^2(t)\right)' = 2 \,B(t)\,\left(B(t)\right)' \quad\iff\quad \cfrac{d \,\left(B\right)^2}{dt} = 2 \,B\,\cfrac{d \, B}{dt}$$

If $B(t) = B(t\,;\, x,y,z)$ is a multi-variate function, the above remains true when the differentiation is taken in $t$ alone while considering the other $x,y,z$ as constant vs. $t$, which is precisely the definition of partial derivatives. Then $\cfrac{d}{dt} \mapsto \cfrac{\partial}{\partial t}$ gives the equality as stated in the original post.