Basic Analysis I — Real Analysis · Problem 4.4.2

The Derivative — Jiri Lebl

Mathematics — Real Analysis · inverse-derivative-chain-rule

Problem

Suppose I,JI, J are intervals and a monotone onto f:I→Jf : I \to J has an inverse g:J→Ig : J \to I. Suppose you already know that both ff and gg are differentiable everywhere and f′f' is never zero. Using chain rule but not Lemma 4.4.1, prove the formula g′(y)=1f′(g(y))g'(y) = \dfrac{1}{f'(g(y))}. Remark: This exercise is the same as Exercise 4.1.10, no need to do it again if you have solved it already.
Intuitive explanation

The key is to use the fundamental relationship between a function and its inverse: f(g(y))=yf(g(y)) = y for all y∈Jy \in J. Differentiate both sides with respect to yy using the chain rule, which gives you a direct equation involving g′(y)g'(y) and f′(g(y))f'(g(y)). Since f′f' is never zero, you can solve for g′(y)g'(y) algebraically—no need to invoke any lemma about inverse derivatives.

Intuitive explanation

The key is to use the fundamental relationship between a function and its inverse: f(g(y))=yf(g(y)) = y for all y∈Jy \in J. Differentiate both sides with respect to yy using the chain rule, which gives you a direct equation involving g′(y)g'(y) and f′(g(y))f'(g(y)). Since f′f' is never zero, you can solve for g′(y)g'(y) algebraically—no need to invoke any lemma about inverse derivatives.

Related theory

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