Bourbaki

Proposition 6

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

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

Let ff be a homeomorphism of an interval I⊂RI \subset \mathbf{R} onto an interval J=f(I)⊂RJ = f(I) \subset \mathbf{R}, and let gg be the inverse homeomorphism3^3. If ff is differentiable at the point x0∈Ix_0 \in I, and if f′(x0)≠0f'(x_0) \neq 0, then gg has a derivative equal to 1/f′(x0)1/f'(x_0) at y0=f(x0)y_0 = f(x_0).

For each y∈Jy \in J we have g(y)∈Ig(y) \in I and u=f(g(y))u = f(g(y)); we thus can write

g(y)−g(y0)y−y0=g(y)−x0f(g(y))−f(x0) \frac{g(y) - g(y_0)}{y - y_0} = \frac{g(y) - x_0}{f(g(y)) - f(x_0)}
for y≠y0y \neq y_0. When yy tends to y0y_0 while remaining in JJ and ≠y0\neq y_0, then g(y)g(y) tends to x0x_0 remaining in II and ≠x0\neq x_0, and the right-hand side in the preceding formula thus has limit 1/f′(x0)1/f'(x_0), since by hypothesis f′(x0)≠0f'(x_0) \neq 0.

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

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Cites

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Transcribed from the printed volume. Not translated.