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.
