# Differentiating an Inner Product

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

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?

As $h\to 0$ the first expression converges to
by definition of the derivative, by continuity of $g$ and by continuity of the scalar product. Hence the desired equality follows.