The Timpson Lyceum · Real Analysis (2020)

Lecture 14 (Real Analysis 2020)

Overview

This lecture introduces two major differentiation rules: the derivative of an inverse function and the chain rule (the derivative of a composition of two functions). Both are proven rigorously from the limit definition of the derivative, with the inverse-function rule resting on the fact that a strictly monotone continuous function has a continuous inverse. The conceptual heart of the lecture is staying disciplined about which variable belongs to which function — confusing where you evaluate $f^{-1}{}'$ is the single most common mistake. The payoff is the general power rule: combining the inverse rule (for $x^{1/n}$ ) with the chain rule (for compositions) extends $\frac{d}{dx}x^n = n x^{n-1}$ from integer exponents to all rational exponents.

Key Concepts

Geometric intuition for the inverse derivative. A derivative is the slope of a function at a point. The inverse function swaps the roles of $x$ and $y$ (reflecting the graph, or equivalently swapping the axes). At the same geometric point, the slope $\frac{\Delta y}{\Delta x}$ becomes $\frac{\Delta x}{\Delta y}$ — the reciprocal. So if $f'(x_0) = m$ , then $(f^{-1})'(y_0) = \frac{1}{m}$ .

The biggest gotcha — where you evaluate. The inverse derivative is taken at $y_0 = f(x_0)$ , not at $x_0$ :

(f^{-1})'(y_0) \;=\; \frac{1}{f'(x_0)}, \qquad y_0 = f(x_0).

It is wrong to write $(f^{-1})'(x_0) = \frac{1}{f'(x_0)}$ . Plugging $x_0$ into $f$ gives $y_0$ ; to recover $x_0$ you must plug $y_0$ into $f^{-1}$ . Keeping one variable tied to each function (write $x = f^{-1}(y)$ and $y = f(x)$ , never re-using $x$ for both) prevents the confusion.

Inverse Function Theorem (derivative form). Let $I$ be an open interval containing $x_0$ and let $f : I \to \mathbb{R}$ be strictly monotone and continuous. Suppose $f'(x_0)$ exists and $f'(x_0) \neq 0$ . Then $f^{-1}$ is differentiable at $y_0 = f(x_0)$ and

(f^{-1})'(y_0) \;=\; \frac{1}{f'(x_0)}.

Strict monotonicity guarantees $f^{-1}$ is continuous (a result proven earlier), which lets the limit pass from $y \to y_0$ to $x \to x_0$ .
The requirement $f'(x_0) \neq 0$ is essential: if the original slope is $0$ , the reciprocal (the inverse's slope) is undefined.
Corollary (interval form). If $f : I \to \mathbb{R}$ is differentiable, strictly monotone, and $f' \neq 0$ everywhere, then $f^{-1}$ is differentiable on the whole image, and $(f^{-1})'(x_0) = \dfrac{1}{f'(f^{-1}(x_0))}$ — the same statement with the variables relabeled.

Chain rule. For the composition $g \circ f$ , the derivative is the derivative of the outside (evaluated at the inside) times the derivative of the inside:

(g \circ f)'(x_0) \;=\; g'\big(f(x_0)\big)\cdot f'(x_0).

The name "chain" reflects that a longer tower of nested functions produces a product chain of derivatives.

Auxiliary-function technicality in the chain-rule proof. Writing $\dfrac{g(y)-g(y_0)}{y-y_0}$ with $y = f(x)$ risks dividing by zero, because without strict monotonicity $y$ can equal $y_0$ at points other than $x_0$ . Define

h(y) = \begin{cases} \dfrac{g(y)-g(y_0)}{y-y_0}, & y \neq y_0,\\ g'(y_0), & y = y_0, \end{cases}

which "fills the hole" at $y_0$ without changing any limit (the limit at a removable hole is unaffected by the value there). Then $\lim_{y\to y_0} h(y) = g'(y_0)$ holds in both cases, and the proof goes through without ever dividing by zero.

General power rule (rational exponents). For $f(x) = x^r$ on $(0,\infty)$ with $r \in \mathbb{Q}$ :

\frac{d}{dx}\, x^{r} \;=\; r\, x^{\,r-1}.

Built in stages: integer exponents (earlier), then $x^{1/n}$ via the inverse rule (it is the inverse of $x^n$ ), then any $r = m/n$ by writing $x^{m/n} = \big(x^{1/n}\big)^{m}$ and applying the chain rule. Writing $r = \tfrac{m}{n}$ with $n \in \mathbb{N}$ keeps the root's domain non-negative.

Worked Example

Verify the inverse-derivative rule for $f(x) = x^2$ on $(0,\infty)$ at $x_0 = 3$ , so $y_0 = f(3) = 9$ . Here $f'(x) = 2x$ , giving $f'(3) = 6$ , so we expect $(f^{-1})'(9) = \tfrac{1}{6}$ . The inverse is $f^{-1}(y) = \sqrt{y}$ ; compute its derivative at $9$ directly from the definition, rationalizing the difference quotient:

\begin{aligned} (f^{-1})'(y_0) &= \lim_{y \to y_0} \frac{f^{-1}(y) - f^{-1}(y_0)}{y - y_0} \\ (f^{-1})'(9) &= \lim_{y \to 9} \frac{\sqrt{y} - 3}{y - 9} = \lim_{y \to 9} \frac{\sqrt{y} - 3}{(\sqrt{y}-3)(\sqrt{y}+3)} \\ &= \lim_{y \to 9} \frac{1}{\sqrt{y} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{6}. \end{aligned}

This matches $\tfrac{1}{f'(3)} = \tfrac{1}{6}$ . Note the evaluation point is $y_0 = 9$ , not $x_0 = 3$ — evaluating the inverse derivative at $3$ would be the classic error.

Summary

The inverse derivative is the reciprocal of the original derivative, evaluated at the matching point: $(f^{-1})'(y_0) = \dfrac{1}{f'(x_0)}$ with $y_0 = f(x_0)$ .
Always evaluate the inverse derivative at $y_0$ , not $x_0$ — keep each variable bound to one function to avoid the most common slip.
The Inverse Function Theorem requires $f$ strictly monotone and continuous with $f'(x_0) \neq 0$ ; monotonicity gives a continuous inverse, and a nonzero slope keeps the reciprocal defined.
The chain rule states $(g \circ f)'(x_0) = g'(f(x_0))\, f'(x_0)$ — derivative of the outside (at the inside) times derivative of the inside.
The chain-rule proof uses an auxiliary function $h(y)$ to fill the removable hole at $y_0$ , avoiding division by zero when $f$ is not monotone.
Combining the inverse rule (for $x^{1/n}$ ) with the chain rule (for $x^{m/n}$ ) extends the power rule to all rational exponents: $\dfrac{d}{dx} x^{r} = r\, x^{r-1}$ .