Bourbaki

Definition 2

Tag 082S, permanent. Cite it as [Bourbaki, Tag 082S].

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

Let f\mathbf{f} be a vector function defined on an interval I⊂RI \subset \mathbf{R}, and let x0x_0 be a point of II such that the interval I∩[x0,+∞[I \cap [x_0, +\infty[ (resp. I∩]−∞,x0]I \cap ]-\infty, x_0]) does not reduce to a single point. We say that f\mathbf{f} is differentiable on the right (resp. on the left) at the point x0x_0 if the restriction of f\mathbf{f} to the interval I∩[x0,+∞[I \cap [x_0, +\infty[ (resp. I∩]−∞,x0]I \cap ]-\infty, x_0]) is differentiable at the point x0x_0; the value of the derivative of this restriction at the point x0x_0 is called the right (resp. left) derivative of f\mathbf{f} at the point x0x_0 and is denoted by fd′(x0)\mathbf{f}'_d(x_0) (resp. fg′(x0)\mathbf{f}'_g(x_0)).

Let f\mathbf{f} be a vector function defined on II, and x0x_0 an interior point of II such that f\mathbf{f} is continuous at this point; it follows from defs. 1 and 2 that for f\mathbf{f} to be differentiable at x0x_0 it is necessary and sufficient that f\mathbf{f} admit both a right and a left derivative at this point, and that these derivatives be equal; and then

f′(x0)=fd′(x0)=fg′(x0). \mathbf{f}'(x_0) = \mathbf{f}'_d(x_0) = \mathbf{f}'_g(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.