If $(V, \langle \cdot, \cdot \rangle)$ is a finite-dimensional inner product space and $f,g : \mathbb{R} \longrightarrow V$ are differentiable functions, a straightforward calculation with components shows that

$$ \frac{d}{dt} \langle f, g \rangle = \langle f(t), g^{\prime}(t) \rangle + \langle f^{\prime}(t), g(t) \rangle $$

This approach is not very satisfying. However, attempting to apply the definition of the derivative directly doesn't seem to work for me. Is there a slick, perhaps intrinsic way, to prove this that doesn't involve working in coordinates?


Observe that $$ \begin{align*} \frac{1}{h} & \left[ \langle f(t+h),\, g(t+h)\rangle - \langle f(t),\, g(t) \rangle \right] \\ & = \frac{1}{h} \left[ \langle f(t+h),\, g(t+h)\rangle - \langle f(t),\, g(t+h)\rangle \right] + \frac{1}{h} \left[ \langle f(t),\, g(t+h)\rangle - \langle f(t),\, g(t)\rangle \right] \\ &= \left\langle \frac{1}{h} \left[ f(t+h) - f(t) \right],\, g(t+h) \right\rangle + \left\langle f(t),\, \frac{1}{h} \left[ g(t+h) - g(t) \right] \right\rangle. \end{align*} $$ As $h\to 0$ the first expression converges to $$ \frac{d}{dt} \langle f(t), g(t) \rangle $$ and the last expression converges to $$ \langle f^{\prime}(t), g(t) \rangle + \langle f(t), g^{\prime}(t) \rangle $$ by definition of the derivative, by continuity of $g$ and by continuity of the scalar product. Hence the desired equality follows.

Note that this doesn't use finite-dimensionality and that the argument is the exact same as the one for the ordinary product rule from calculus.


This answer may be needlessly complicated if you don't want such generality, taking the approach of first finding the Fréchet derivative of a bilinear operator.

If $V$, $W$, and $Z$ are normed spaces, and if $T:V\times W\to Z$ is a continuous (real) bilinear operator, meaning that there exists $C\geq 0$ such that $\|T(v,w)\|\leq C\|v\|\|w\|$ for all $v\in V$ and $w\in W$, then the derivative of $T$ at $(v_0,w_0)$ is $DT|_{(v_0,w_0)}(v,w)=T(v,w_0)+T(v_0,w)$. (I am assuming that $V\times W$ is given a norm equivalent with $\|(v,w)\|=\sqrt{\|v\|^2+\|w\|^2}$.) This follows from the straightforward computation

$$\frac{\|T(v_0+v,w_0+w)-T(v_0,w_0)-(T(v,w_0)+T(v_0,w))\|}{\|(v,w)\|}=\frac{\|T(v,w)\|}{\|(v,w)\|}\leq C\frac{\|v\|\|w\|}{\|(v,w)\|}\to 0$$

as $(v,w)\to 0$.

With $V=W$, $Z=\mathbb R$ or $Z=\mathbb C$, and $T:V\times V\to Z$ the inner product, this gives $DT_{(v_0,w_0)}(v,w)=\langle v,w_0\rangle+\langle v_0,w\rangle$. Now if $f,g:\mathbb R\to V$ are differentiable, then $F:\mathbb R\to V\times V$ defined by $F(t)=(f(t),g(t))$ is differentiable with $DF|_t(h)=h(f'(t),g'(t))$. By the chain rule,

$$D(T\circ F)|_{t}(h) =DT|_{F(t)}\circ DF|_t(h)=h(\langle f'(t),g(t)\rangle+\langle f(t),g'(t)\rangle),$$

which means $\frac{d}{dt} \langle f, g \rangle = \langle f'(t),g(t)\rangle+\langle f(t),g'(t)\rangle$.

  • 3
    $\begingroup$ I notice that $\langle v, w_0\rangle + \langle v_0, w\rangle$ is not linear in $(v,w)$. (I.e. it's conjugate linear in $w$). So how can this be the derivative? I stumbled upon this page with exactly this question in mind. $\endgroup$ – Eric Auld Jan 16 '15 at 1:02
  • 3
    $\begingroup$ @Eric: It is real linear. That it what is used here, as noted above, "if $T$... is a continuous (real) bilinear operator...". $\endgroup$ – Jonas Meyer Jan 16 '15 at 2:33
  • $\begingroup$ This answer also generalizes nicely to multilinear functions too. $\endgroup$ – Vasting Dec 29 '20 at 22:56

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