We start with a simple example. Consider the function \(f(x) \coloneqq a x\) for a number \(a \neq 0\text{.}\) Then \(f \colon \R \to \R\) is bijective, and the inverse is \(f^{-1}(y) = \frac{1}{a} y\text{.}\) In particular, \(f'(x) = a\) and \((f^{-1})'(y) = \frac{1}{a}\text{.}\) As differentiable functions are “infinitesimally like” linear functions, we expect the same sort of behavior from the inverse of a differentiable function. The main idea of differentiating inverse functions is the following lemma.
Let \(I,J \subset \R\) be intervals. If \(f \colon I \to J\) is strictly monotone (hence one-to-one), onto (\(f(I) = J\)), differentiable at \(x_0 \in I\text{,}\) and \(f'(x_0) \neq 0\text{,}\) then the inverse \(f^{-1}\) is differentiable at \(y_0 = f(x_0)\) and
By Proposition 3.6.6, \(f\) has a continuous inverse. For convenience, call the inverse \(g \colon J \to I\text{.}\) Let \(x_0,y_0\) be as in the statement. For \(x \in I\text{,}\) write \(y \coloneqq f(x)\text{.}\) If \(x \neq x_0\text{,}\) and so \(y \neq y_0\text{,}\) we find
that is, \(Q\) is continuous at \(x_0\text{.}\) As \(g(y)\) is continuous at \(y_0\text{,}\) the composition \(Q\bigl(g(y)\bigr) = \frac{g(y)-g(y_0)}{y-y_0}\) is continuous at \(y_0\) by Proposition 3.2.7. Therefore,
If \(f'\) is continuous and nonzero at all \(x \in I\text{,}\) then the lemma applies at all \(x \in I\text{.}\) As \(g\) is also continuous (it is differentiable), the derivative \(g'(y) =
\frac{1}{f'\left(\vphantom{1^1_1}g(y)\right)}\) must be continuous.
Let \(f \colon (a,b) \to \R\) be a continuously differentiable function, \(x_0 \in (a,b)\) a point where \(f'(x_0) \neq 0\text{.}\) Then there exists an open interval \(I \subset (a,b)\) with \(x_0 \in I\text{,}\) the restriction \(f|_{I}\) is injective with a continuously differentiable inverse \(g \colon J \to I\) defined on an interval \(J \coloneqq f(I)\text{,}\) and
\begin{equation}
g'(y) = \frac{1}{f'\bigl( g(y) \bigr)} \qquad \text{for all } y \in J.
\end{equation}
Without loss of generality, suppose \(f'(x_0) > 0\text{.}\) As \(f'\) is continuous, there must exist an open interval \(I = (x_0-\delta,x_0+\delta)\) such that \(f'(x) > 0\) for all \(x \in I\text{.}\) See Exercise 3.2.11.
By Proposition 4.2.8, \(f\) is strictly increasing on \(I\text{,}\) and hence the restriction \(f|_{I}\) is bijective onto \(J \coloneqq f(I)\text{.}\) As \(f\) is continuous, Corollary 3.6.3 (or directly via the intermediate value theorem) implies that \(f(I)\) is an interval. Now apply Lemma 4.4.1.
In Example 1.2.3, we saw how difficult an endeavor was proving the existence of \(\sqrt{2}\) without any tools. With the intermediate value theorem, the existence of roots is almost trivial, and with the machinery of this section, we will prove far more than mere existence.
Given \(n \in \N\) and \(x \geq 0\text{,}\) there exists a unique number \(y \geq 0\) (denoted \(x^{1/n} \coloneqq y\)), such that \(y^n = x\text{.}\) Furthermore, the function \(g \colon (0,\infty) \to (0,\infty)\) defined by \(g(x) \coloneqq x^{1/n}\) is continuously differentiable and
Let \(f \colon (0,\infty) \to (0,\infty)\) be defined by \(f(y) \coloneqq y^n\text{.}\) The function \(f\) is continuously differentiable, and \(f'(y) = ny^{n-1}\text{,}\) see Exercise 4.1.3. For \(y > 0\text{,}\) the derivative \(f'\) is strictly positive, and so again by Proposition 4.2.8, \(f\) is strictly increasing (this can also be proved directly) and hence injective. Suppose \(M\) and \(\epsilon\) are such that \(M > 1\) and \(1 > \epsilon > 0\text{.}\) Then \(f(M) = M^n \geq M\) and \(f(\epsilon) = \epsilon^n \leq \epsilon\text{.}\) For every \(x\) with \(\epsilon < x < M\text{,}\) we have, by the intermediate value theorem, that \(x \in
f\bigl( [\epsilon,M] \bigr) \subset
f\bigl( (0,\infty) \bigr)\text{.}\) As \(M\) and \(\epsilon\) were arbitrary, \(f\) is onto \((0,\infty)\text{,}\) and hence \(f\) is bijective. Let \(g\) be the inverse of \(f\text{,}\) and we obtain the existence and uniqueness of positive \(n\)th roots. Lemma 4.4.1 says that \(g\) has a continuous derivative and \(g'(x) =
\frac{1}{f'\left(\vphantom{1^1_1}g(x)\right)} = \frac{1}{n {(x^{1/n})}^{n-1}}\text{.}\)
The corollary provides a good example of where the inverse function theorem gives us an interval smaller than \((a,b)\text{.}\) Take \(f \colon \R \to \R\) defined by \(f(x) \coloneqq x^2\text{.}\) Then \(f'(x_0) \neq 0\) as long as \(x_0 \neq 0\text{.}\) If \(x_0 > 0\text{,}\) we can take \(I=(0,\infty)\text{,}\) but no larger.
Another useful example is \(f(x) \coloneqq x^3\text{.}\) The function \(f \colon \R \to \R\) is one-to-one and onto, so \(f^{-1}(y) = y^{1/3}\) exists on the entire real line, including zero and negative \(y\text{.}\) The function \(f\) has a continuous derivative, but \(f^{-1}\) has no derivative at the origin. The point is that \(f'(0) = 0\text{.}\) See Figure 4.11 for a graph. Notice the vertical tangent on the cube root at the origin. See also Exercise 4.4.4.
Suppose \(f \colon \R \to \R\) is continuously differentiable and \(f'(x) > 0\) for all \(x\text{.}\) Show that \(f\) is invertible on the interval \(J =
f(\R)\text{,}\) the inverse is continuously differentiable, and \({(f^{-1})}'(y) >
0\) for all \(y \in f(\R)\text{.}\)
Suppose \(I,J\) are intervals and a monotone onto \(f \colon I \to J\) has an inverse \(g \colon J \to I\text{.}\) Suppose you already know that both \(f\) and \(g\) are differentiable everywhere and \(f'\) is never zero. Using the chain rule but not Lemma 4.4.1, prove the formula \(g'(y) = \frac{1}{f'\left(\vphantom{1_1^1}g(y)\right)}\text{.}\) Remark: This exercise is the same as Exercise 4.1.10, no need to do it again if you have solved it already.
Let \(n\in \N\) be even. Prove that every \(x > 0\) has a unique negative \(n\)th root. That is, there exists a negative number \(y\) such that \(y^n = x\text{.}\) Compute the derivative of the function \(g(x) \coloneqq y\text{.}\)
Let \(n \in \N\) be odd and \(n \geq 3\text{.}\) Prove that every \(x\) has a unique \(n\)th root. That is, there exists a number \(y\) such that \(y^n = x\text{.}\) Prove that the function defined by \(g(x) \coloneqq y\) is differentiable except at \(x=0\) and compute the derivative. Prove that \(g\) is not differentiable at \(x=0\text{.}\)
(requires Section 4.3) Show that if in the inverse function theorem \(f\) has \(k\) continuous derivatives, then the inverse function \(g\) also has \(k\) continuous derivatives.
Let \(f(x) \coloneqq x + 2 x^2 \sin(\nicefrac{1}{x})\) for \(x \neq 0\) and \(f(0) \coloneqq 0\text{.}\) Show that \(f\) is differentiable at all \(x\text{,}\) that \(f'(0) > 0\text{,}\) but that \(f\) is not invertible on any open interval containing the origin.
Let \(f \colon \R \to \R\) be a continuously differentiable function and \(k > 0\) be a number such that \(f'(x) \geq k\) for all \(x \in \R\text{.}\) Show \(f\) is one-to-one and onto, and has a continuously differentiable inverse \(f^{-1} \colon \R \to \R\text{.}\)
Suppose \(I,J\) are intervals and a monotone onto \(f \colon I \to J\) has an inverse \(g \colon J \to I\text{.}\) Suppose \(x \in I\) and \(y \coloneqq f(x) \in J\text{,}\) and that \(g\) is differentiable at \(y\text{.}\) Prove:
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