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Derivative of Inverse Functions

MATH162
Reference: Stewart 6.1  •  Chapter: 6  •  Section: 1

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Derivative of Inverse Functions

The Reciprocal Relationship

If you know how fast $f$ changes, can you figure out how fast $f^{-1}$ changes? Yes—and the answer is elegant: the slopes are reciprocals.

Think geometrically: if the graph of $f^{-1}$ is the reflection of $f$ across the line $y = x$, then tangent lines also get reflected. Reflection swaps the rise and run, turning slope $m$ into slope $1/m$.

Before You Start: Quick Self-Check

Can you answer these questions?

  1. If $f(5) = 8$, what is $f^{-1}(8)$? Answer: 5 (by definition of inverse).
  2. What is $\frac{d}{dx}[x^3]$? Answer: $3x^2$ (power rule).
  3. What does the Chain Rule say? Answer: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$.
If you struggled with these, review first

This skill requires:

Prerequisite Map

Prerequisites
Finding Inverse FunctionsDerivative DefinitionChain RuleImplicit Differentiation
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This skill
Derivative of Inverse Functions
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Unlocks
Derivative of ln xDerivatives of Inverse TrigDerivatives of Inverse Hyperbolic

Quick Reference

Property Value
Chapter Chapter 6: Inverse Functions
Section §6.1 Inverse Functions and Their Derivatives
Difficulty Intermediate
Time ~20 minutes

Key Concepts

The Derivative of Inverse Formula

If $f$ is a one-to-one differentiable function with inverse $f^{-1}$, and $f'(f^{-1}(a)) \neq 0$, then:

$$\boxed{(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}}$$

In words: The derivative of $f^{-1}$ at $a$ is the reciprocal of the derivative of $f$ at the corresponding point.

Leibniz Notation

If $y = f^{-1}(x)$, then $x = f(y)$, and:

$$\frac{dy}{dx} = \frac{1}{dx/dy}$$

This says: the rate of change of $y$ with respect to $x$ is the reciprocal of the rate of change of $x$ with respect to $y$.

Why This Works: Geometric Intuition

        y                           y
        |      tangent to f         |     tangent to f⁻¹
        |         /                 |        __/
        |        /  slope = m       |      _/   slope = 1/m  
        |    ___/                   |    _/
        |   /                       |  _/
        |  * (b, a)                 | * (a, b)
        |________________________ x |________________________ x

     Graph of f                  Graph of f⁻¹ (reflected)

At $(b, a)$ on $f$: slope = $f'(b) = m$

At $(a, b)$ on $f^{-1}$: slope = $(f^{-1})'(a) = 1/m$

Reflecting swaps rise and run, so slope becomes reciprocal.

The Formula in Action

Setting: Want $(f^{-1})'(a)$.

Step 1: Find $b = f^{-1}(a)$, meaning find $b$ such that $f(b) = a$.

Step 2: Compute $f'(b)$.

Step 3: Take the reciprocal: $(f^{-1})'(a) = \frac{1}{f'(b)}$.

Proof via Implicit Differentiation

If $y = f^{-1}(x)$, then $f(y) = x$.

Differentiate both sides with respect to $x$: $$f'(y) \cdot \frac{dy}{dx} = 1$$

Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{1}{f'(y)} = \frac{1}{f'(f^{-1}(x))}$$

Practice Problems

Level 1 State the Formula

Complete the formula: If $f(5) = 8$ and $f'(5) = 3$, then $(f^{-1})'(8) = $ .

Thought Process

Goal: Find $(f^{-1})'(8)$ given information about $f$.

The formula: $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$

Step 1: Identify $a$. Here $a = 8$.

Step 2: Find $f^{-1}(8)$.

  • We know $f(5) = 8$.
  • So $f^{-1}(8) = 5$.

Step 3: Plug into the formula.

  • $(f^{-1})'(8) = \frac{1}{f'(f^{-1}(8))} = \frac{1}{f'(5)} = \frac{1}{3}$
Show Answer

$(f^{-1})'(8) = \frac{1}{f'(5)} = \boxed{\frac{1}{3}}$

Level 2 Using the Formula Directly

Let $f(x) = x^3 + 1$. Find $(f^{-1})'(9)$ without finding $f^{-1}$ explicitly.

Thought Process

Goal: Find $(f^{-1})'(9)$ for $f(x) = x^3 + 1$.

The 3-step process:

Step 1: Find $f^{-1}(9)$ by solving $f(b) = 9$.

  • $b^3 + 1 = 9$
  • $b^3 = 8$
  • $b = 2$
  • So $f^{-1}(9) = 2$.

Step 2: Compute $f'(x)$.

  • $f'(x) = 3x^2$

Step 3: Evaluate at the point from Step 1 and take reciprocal.

  • $f'(2) = 3(2)^2 = 12$
  • $(f^{-1})'(9) = \frac{1}{12}$
Show Answer

Step 1: Find $f^{-1}(9)$: solve $x^3 + 1 = 9$, so $x^3 = 8$, get $x = 2$.

Step 2: Compute $f'(2) = 3(2)^2 = 12$.

Step 3: $(f^{-1})'(9) = \frac{1}{f'(2)} = \boxed{\frac{1}{12}}$

Verification: We have $f^{-1}(x) = (x-1)^{1/3}$, so $(f^{-1})'(x) = \frac{1}{3}(x-1)^{-2/3}$.

At $x = 9$: $(f^{-1})'(9) = \frac{1}{3} \cdot 8^{-2/3} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$ ✓

Level 3 Finding the Right Point

Let $f(x) = x^2 + 3x$ for $x \geq 0$. Find $(f^{-1})'(10)$.

Thought Process

First, find $b$ such that $f(b) = 10$, i.e., $b^2 + 3b = 10$.

Solving: $b^2 + 3b - 10 = 0$, so $(b+5)(b-2) = 0$, giving $b = 2$ or $b = -5$.

Since domain is $x \geq 0$, take $b = 2$.

Then $f'(x) = 2x + 3$, so $f'(2) = 7$.

Show Answer

Step 1: Find $f^{-1}(10)$ by solving $x^2 + 3x = 10$: $$x^2 + 3x - 10 = 0$$ $$(x+5)(x-2) = 0$$

Since $x \geq 0$, we get $x = 2$.

Step 2: Compute $f'(2) = 2(2) + 3 = 7$.

Step 3: $(f^{-1})'(10) = \frac{1}{f'(2)} = \boxed{\frac{1}{7}}$

Level 4 Transcendental Function

Let $f(x) = 2x + \cos x$. Find $(f^{-1})'(1)$.

Note: $f$ is one-to-one because $f'(x) = 2 - \sin x > 0$ for all $x$.

Thought Process

I need to find $b$ such that $f(b) = 1$, i.e., $2b + \cos b = 1$.

By inspection: $b = 0$ gives $2(0) + \cos(0) = 0 + 1 = 1$. ✓

So $f^{-1}(1) = 0$.

Then $f'(0) = 2 - \sin 0 = 2 - 0 = 2$.

Show Answer

Step 1: Find $f^{-1}(1)$ by solving $2x + \cos x = 1$.

By inspection: $f(0) = 2(0) + \cos(0) = 1$. ✓

So $f^{-1}(1) = 0$.

Step 2: Compute $f'(0) = 2 - \sin 0 = 2$.

Step 3: $(f^{-1})'(1) = \frac{1}{f'(0)} = \boxed{\frac{1}{2}}$

Level 5 Deriving the Formula

Prove the derivative of inverse formula using the definition of derivative:

$$(f^{-1})'(a) = \lim_{x \to a} \frac{f^{-1}(x) - f^{-1}(a)}{x - a}$$

Thought Process

Let $y = f^{-1}(x)$ and $b = f^{-1}(a)$, so $f(y) = x$ and $f(b) = a$.

Then as $x \to a$, we have $y \to b$ (since $f^{-1}$ is continuous).

Rewrite the difference quotient in terms of $y$ and $b$.

Show Answer

Proof:

Let $b = f^{-1}(a)$, so $f(b) = a$.

Let $y = f^{-1}(x)$, so $f(y) = x$.

Since $f^{-1}$ is continuous (as inverse of continuous function), $y \to b$ as $x \to a$.

Now: $$(f^{-1})'(a) = \lim_{x \to a} \frac{f^{-1}(x) - f^{-1}(a)}{x - a} = \lim_{y \to b} \frac{y - b}{f(y) - f(b)}$$

$$= \lim_{y \to b} \frac{1}{\frac{f(y) - f(b)}{y - b}} = \frac{1}{\lim_{y \to b} \frac{f(y) - f(b)}{y - b}} = \frac{1}{f'(b)} = \frac{1}{f'(f^{-1}(a))}$$

$\square$

CCI-Style Conceptual Questions

Level 2 Conceptual: When Does the Formula Fail?

At a point where $f'(b) = 0$, what can you say about $(f^{-1})'(a)$ where $a = f(b)$?

Thought Process

The formula says $(f^{-1})'(a) = 1/f'(b)$. If $f'(b) = 0$, we're dividing by zero.

Geometrically: if $f$ has a horizontal tangent at $(b, a)$, then $f^{-1}$ has a vertical tangent at $(a, b)$.

Show Answer

If $f'(b) = 0$, then $(f^{-1})'(a)$ is undefined (the formula gives $1/0$).

Geometrically: A horizontal tangent to $f$ reflects to a vertical tangent to $f^{-1}$. The inverse function is not differentiable at that point.

Example: $f(x) = x^2$ on $[0,\infty)$ has $f'(0) = 0$. Its inverse $f^{-1}(x) = \sqrt{x}$ has derivative $(f^{-1})'(x) = \frac{1}{2\sqrt{x}}$, which is undefined at $x = 0$ (vertical tangent).

Mastery Checklist

Mental Model

The "Reciprocal Slopes" Picture:

Imagine the tangent line to $f$ at point $(b, a)$. It has some slope $m = f'(b)$.

Now reflect everything about $y = x$. The point becomes $(a, b)$, and the tangent line's rise and run swap. A slope of $m$ becomes a slope of $1/m$.

That reflected tangent is the tangent to $f^{-1}$ at $(a, b)$, so $(f^{-1})'(a) = 1/m = 1/f'(b)$.

Key formula to memorize: $$\text{slope of } f^{-1} \text{ at output} = \frac{1}{\text{slope of } f \text{ at corresponding input}}$$

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Last updated: 2026-01-22