Differentiating an Inner Product

If (V,,) is a finite-dimensional inner product space and f,g:RV are differentiable functions, a straightforward calculation with components shows that

ddtf,g=f(t),g(t)+f(t),g(t)

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?

Answer

Observe that
1h[f(t+h),g(t+h)f(t),g(t)]=1h[f(t+h),g(t+h)f(t),g(t+h)]+1h[f(t),g(t+h)f(t),g(t)]=1h[f(t+h)f(t)],g(t+h)+f(t),1h[g(t+h)g(t)].
As h0 the first expression converges to
ddtf(t),g(t) and the last expression converges to
f(t),g(t)+f(t),g(t)
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.

Attribution
Source : Link , Question Author : ItsNotObvious , Answer Author : t.b.

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