Bourbaki

Example 3

Tag 083B, permanent. Cite it as [Bourbaki, Tag 083B].

Functions of a Real Variable, chapter I, § 1. FIRST DERIVATIVE

The euclidean scalar product (x∣y)(\mathbf{x}| \mathbf{y}) (Gen. Top , VI, p. 40) is a bilinear map (necessarily continuous) of Rn×Rn\mathbf{R}^n \times \mathbf{R}^n into R\mathbf{R}. If f\mathbf{f} and g\mathbf{g} are two vector functions with values in Rn\mathbf{R}^n, and differentiable at the point x0x_0, then the real function x↦(f(x)∣g(x))x \mapsto (\mathbf{f}(x)|\mathbf{g}(x)) has a derivative equal to (f′(x0)∣g(x0))+(f(x0)∣g′(x0))(\mathbf{f}'(x_0)|\mathbf{g}(x_0)) + (\mathbf{f}(x_0)|\mathbf{g}'(x_0)) at the point x0x_0. There is an analogous result for the hermitian scalar product on Cn\mathbf{C}^n, this space being considered as a vector space over R\mathbf{R}.

Let us consider in particular the case where the euclidean norm ∥f(x)∥\| \mathbf{f}(x) \| is constant, so that (f(x)∣f(x))=∥f(x)∥2(\mathbf{f}(x)|\mathbf{f}(x)) = \| \mathbf{f}(x) \|^2 is also constant; on writing that the derivative of (f(x)∣f(x))(\mathbf{f}(x)|\mathbf{f}(x)) vanishes at x0x_0 we obtain (f(x0)∣f′(x0))=0(\mathbf{f}(x_0)|\mathbf{f}'(x_0)) = 0; in other words, f′(x0)\mathbf{f}'(x_0) is orthogonal to f(x0)\mathbf{f}(x_0).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.