Product rules

If $f(t)$ and $g(t)$ are scalar functions, we know that $\frac{d}{dt} [f(t)g(t)] = f'(t)g(t) + f(t) g'(t)$. But what about vector-valued functions ${\bf u}(t)$ and ${\bf v(t)}$?

The product rules for dot products and cross products are almost the same as the product rules for scalar functions: $\displaystyle{\frac{d}{dt} [f(t){\bf u}(t)] = f'(t) {\bf u}(t) + f(t) {\bf u}'(t)}.$ $\displaystyle{\frac{d}{dt} [{\bf u}(t)\cdot {\bf v}(t)] = {\bf u}'(t)\cdot {\bf v}(t) + {\bf u}(t)\cdot {\bf v}'(t).}$ $\displaystyle{\frac{d}{dt} [{\bf u}(t)\times {\bf v}(t)] = {\bf u}'(t)\times {\bf v}(t) + {\bf u}(t)\times {\bf v}'(t).}$

Be careful with the order in the last equation! We want ${\bf u} \times {\bf v}'$, not ${\bf v}' \times {\bf u}$! When applying rules from calculus or algebra to vector products, you always have to preserve the order of the vectors.

The chain rule applies to expressions like ${\bf u}(f(t))$, where $f(t)$ is a scalar function: $$\frac{d}{dt} {\bf u}({f(t)}) = {\bf u}'(f(t)) \, f'(t).$$

These formulas are all proved the same way. We just write everything out in components and apply the usual product rule, or the usual chain rule to each component. For instance, \begin{eqnarray*} \frac{d}{dt} [f(t) {\bf u}(t)] & = & \frac{d}{dt} \langle f(t) u_1(t) , f(t) u_2(t), f(t) u_3(t) \rangle \cr \cr & = & \left \langle \frac{d}{dt} [f(t) u_1(t)], \frac{d}{dt} [f(t) u_2(t)], \frac{d}{dt} [f(t) u_3(t)] \right \rangle \cr \cr & = & \left \langle f'(t) u_1(t) + f(t) u_1'(t), f'(t) u_2(t) + f(t) u_2'(t), f'(t) u_3(t) + f(t) u_3'(t) \right \rangle \cr \cr & = & f'(t) \langle u_1(t), u_2(t), u_3(t) \rangle + f(t) \langle u_1'(t), u_2'(t), u_3'(t) \rangle \cr \cr & = & f'(t) {\bf u}(t) + f(t) {\bf u}'(t). \end{eqnarray*}Likewise, \begin{eqnarray*} \frac{d}{dt} [{\bf u}(t) \cdot {\bf v}(t)] & = & \frac{d}{dt} [u_1(t)v_1(t) + u_2(t)v_2(t) + u_3(t)v_3(t)] \cr \cr & = & u_1'(t) v_1(t) + u_1(t) v_1'(t) + u_2'(t) v_2(t) + u_2(t) v_2'(t) + u_3'(t) v_3(t) + u_3(t) v_3'(t) \cr \cr & = & u_1'(t) v_1(t) + u_2'(t) v_2(t) + u_3'(t) v_3(t) + u_1(t) v_1'(t) + u_2(t) v_2'(t) + u_3(t) v_3'(t) \cr\cr & = & {\bf u}'(t) \cdot {\bf v}(t) + {\bf u}(t) \cdot {\bf v}'(t). \end{eqnarray*}The proof of the cross product identity follows the same pattern, only with more terms to keep track of. The chain rule is similar.