Differentiating Inverses

In this calculus series, we will assume that the functions $\sin$ and $\exp(x) \equiv e^x$ have been well-defined. These allow for definitions of common trigonometric functions like $\cos$ .

We want to differentiate their inverses $\sin^{-1}$ and $\ln$ . Here is the definition of the inverse of a function.

Definition. A function $f$ defined on $I \subseteq \mathbb R$ has an inverse if for any $r, s \in I$ , $f(r) = f(s) \Rightarrow r = s$ .

However, a technical discussion on the properties of $\sin$ and $\exp$ will take us beyond the focus of differentiation. For this discussion, we will make the following assumptions:

$\sin$ restricted to $[-\pi/2,\pi/2]$ has an inverse $\sin^{-1}$ .
$\exp$ has an inverse $\ln$ .
$\sin$ is differentiable on $\mathbb R$ and $\sin' = \cos$ .
$\exp$ is differentiable on $\mathbb R$ and $\exp' = \exp$ .

We will rigorously define these functions more carefully, and prove these properties, when studying the big brother of calculus that grounds it, known as real analysis.

Before we can prove the inverse function theorem, which allows us to differentiate inverses, we first need some technical tools. The first is the famous intermediate value theorem, which can only be fully proven using real-analytic tools, and thus is omitted from this discussion.

Theorem 1 (Intermediate Value Theorem). Let $f : [a, b] \to \mathbb R$ be continuous. Suppose $f(a) \neq f(b)$ without loss of generality. Then for any $f(a) < m < f(b)$ , there exists $c \in (a, b)$ such that $f(c) = m$ .

With some effort, we can derive the general case by first proving the special case $[a, b] = [0, 1]$ , $f(a) < 0 < f(b)$ , and $m = 0$ . We will relegate that in a discussion on real analysis.

Using the intermediate value theorem, we will first show that any interval remains an interval.

Lemma 1. Let $f$ be continuous on $\mathbb R$ . For any $a < b$ , define $I := (a, b)$ . Then for any $c, d \in f(I)$ with $c < d$ , if $c < y < d$ , then $y \in f(I)$ .

Proof. Fix $c, d \in f(I)$ with $c < d$ . Find inputs $r, s \in (a, b)$ such that $f(r) = c < d = f(s)$ . Suppose $r < s$ without loss of generality.

Fix $f(r) < y < f(s)$ . Since $f$ is continuous on $[r, s]$ , by the intermediate value theorem, there exists $t \in (r, s)$ such that $y = f(t) \in f(I)$ .

We will next show that the inverse of any continuous function must be continuous.

Lemma 2. Let $f$ be a real-valued function that is continuous on $(a, b)$ . Suppose $f$ has an inverse $f^{-1}$ . Then $f^{-1}$ is continuous on the interval $f((a, b))$ .

Proof. Fix $c \in f((a, b))$ . By the intermediate value theorem, find $d \in (a, b)$ such that $c = f(d) \iff f^{-1}(c) = d$ . Since $f$ is continuous at $d$ ,

$f^{-1}(c + o(1)) = f^{-1}(f(d) + o(1)) = f^{-1}(f(d + o(1))) = d + o(1) .$

Finally, here is the inverse function theorem in one dimension.

Theorem 2 (Inverse Function Theorem). Let $f$ be differentiable on $(a, b)$ . Suppose $f$ has an inverse $f^{-1}$ on $(a, b)$ . Then for any $c \in f((a, b))$ such that $f'(f(c)) \neq 0$ , $f^{-1}$ is differentiable at $c$ and

$\displaystyle (f^{-1})'(c) = \frac{1}{f'(f^{-1}(c))}.$

The theorem for higher-dimensions takes a lot more effort to define and prove. For this post, we will omit that.

Proof. The key component of this proof is Carathéodory’s theorem, which we have proven previously.

Let $y = f(x) \iff x = f^{-1}(y)$ and $c = f(d) \iff d = f^{-1}(c)$ . Then

$\displaystyle \begin{aligned} f^{-1}(y) - f^{-1}(c) &= \frac{f^{-1}(y) - f^{-1}(c)}{y-c} \cdot (y-c) \\ &= \frac{1}{\frac{y-c}{f^{-1}(y) - f^{-1}(c)}} \cdot (y-c) \\ &= \frac 1{\frac{f(x) - f(d)}{x-d}} \cdot (y-c). \end{aligned}$

By Carathéodory’s theorem, find a continuous function $\varphi$ such that

$f(x) - f(d) = \varphi(x) (x-d)$

with $(\varphi \circ f^{-1})(c) = \varphi(d) = f'(d) = f'(f^{-1}(c))$ .

Substituting,

$\displaystyle \begin{aligned} f^{-1}(y) - f^{-1}(c) &= \frac 1{\frac{f(x) - f(d)}{x-d}} \cdot (y-c) = \frac 1{(\varphi \circ f^{-1}) (x)} (y-c). \end{aligned}$

By Caratheodory’s theorem, define the continuous function $\psi := 1/{(\varphi \circ f^{-1})}$ , which satisfies $(f^{-1})'(c) = \psi(c) = 1/{f'(f^{-1}(c))}$ , to complete the result.

As a consequence, we obtain the differentiability of $\sin^{-1}$ on $(-\pi/2,\pi/2)$ and $\ln = \exp'$ on $(0,\infty)$ . To obtain their derivatives $(\sin^{-1})'(x) = 1/{\sqrt{1-x^2}}$ and $\ln'(x) = 1/x$ , either apply the inverse function theorem, or simply use the chain rule.

To illustrate this technique, let’s prove that $\ln'(x) = 1/x$ .

Theorem 3. $\ln'(x) = 1/x$ .

Proof. By the inverse function theorem,

$\displaystyle \ln'(x) = (\exp^{-1})'(x) = \frac{1}{\exp'(\exp^{-1}(x))} = \frac{1}{\exp(\exp^{-1}(x))} = \frac 1x.$

The derivative of $\ln$ helps us obtain one of the most lovely results in differentiation.

Theorem 3. Let $r$ be a nonzero real number. Then for positive $x$ , $(x^r)' = rx^{r-1}$ .

Proof. Writing $x^r = e^{r \ln(x)} \equiv \exp(r\ln(x))$ and applying the chain rule,

$\displaystyle \begin{aligned} (x^r)' &= (\exp(r\ln(x)))' \\ &= \exp'(r\ln(x)) (r\ln(x))' \\ &= \exp(r \ln(x)) \cdot r \cdot \frac 1x \\ &= x^r \cdot \frac rx = rx^{r-1}. \end{aligned}$

—Joel Kindiak, 21 Oct 24, 1126H

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Differentiating Inverses

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