2.9 – Derivatives of Inverse Functions

Learning Objectives

By the end of this section, you should be able to:

  • Understand the relationship between a function and its inverse
  • Apply the formula for the derivative of an inverse function
  • Find derivatives of inverse trigonometric functions
  • Use implicit differentiation to derive inverse function derivatives
  • Solve problems involving slopes of inverse functions
  • Apply the inverse function theorem to various function types
Section Progress

Understanding Inverse Functions

If \( f \) and \( g \) are inverse functions, then:

  • \( f(g(x)) = x \) for all x in the domain of g
  • \( g(f(x)) = x \) for all x in the domain of f
  • The graphs of f and g are reflections across the line \( y = x \)

Notation: If \( f \) and \( g \) are inverses, we write \( g = f^{-1} \)

The Inverse Function Theorem

If \( f \) is differentiable and has an inverse function \( f^{-1} \), then:

\[ \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))} \]

In words: The derivative of the inverse function is the reciprocal of the derivative of the original function, evaluated at the corresponding point.

Key Insight: Reciprocal Relationship

If \( (a, b) \) is a point on \( f \), then \( (b, a) \) is a point on \( f^{-1} \).

The slopes at these corresponding points are reciprocals:

\[ f'(a) = m \quad \Rightarrow \quad (f^{-1})'(b) = \frac{1}{m} \]

This makes geometric sense because reflection across \( y = x \) transforms slope \( m \) to slope \( \frac{1}{m} \).

Inverse Derivative Calculator

Explore the relationship between a function and its inverse derivative:

Worked Examples

Example 1: Linear Function Inverse

Basic Application

Problem: Let \( f(x) = 2x + 3 \). Find the derivative of \( f^{-1}(x) \) at \( x = 7 \).

1

Find the point on f corresponding to x = 7 on f⁻¹:

Since \( f^{-1}(7) \) is some value a where \( f(a) = 7 \):

\[ 2a + 3 = 7 \Rightarrow 2a = 4 \Rightarrow a = 2 \]

So \( f^{-1}(7) = 2 \)

2

Find f'(x):

\[ f'(x) = 2 \]

3

Apply the inverse function theorem:

\[ (f^{-1})'(7) = \frac{1}{f'(f^{-1}(7))} = \frac{1}{f'(2)} = \frac{1}{2} \]

4

Verification:

Find \( f^{-1}(x) \) explicitly: \( y = 2x + 3 \Rightarrow x = \frac{y - 3}{2} \Rightarrow f^{-1}(x) = \frac{x - 3}{2} \)

\[ (f^{-1})'(x) = \frac{1}{2} \Rightarrow (f^{-1})'(7) = \frac{1}{2} \] ✓

Example 2: Quadratic Function Inverse

Restricted Domain

Problem: Let \( f(x) = x^2 \) for \( x \geq 0 \). Find \( (f^{-1})'(4) \).

1

Find the corresponding point:

We need \( f^{-1}(4) \):

\[ f(a) = 4 \Rightarrow a^2 = 4 \Rightarrow a = 2 \] (since \( x \geq 0 \))

So \( f^{-1}(4) = 2 \)

2

Find f'(x):

\[ f'(x) = 2x \]

3

Apply the theorem:

\[ (f^{-1})'(4) = \frac{1}{f'(f^{-1}(4))} = \frac{1}{f'(2)} = \frac{1}{2 \cdot 2} = \frac{1}{4} \]

4

Verification:

\( f^{-1}(x) = \sqrt{x} \) for \( x \geq 0 \)

\[ (f^{-1})'(x) = \frac{1}{2\sqrt{x}} \Rightarrow (f^{-1})'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4} \] ✓

Example 3: Using Implicit Differentiation

Alternative Method

Problem: Derive the formula for the derivative of \( \ln(x) \) using the fact that it's the inverse of \( e^x \).

1

Set up the inverse relationship:

Let \( y = \ln(x) \). Then by definition of inverse:

\[ e^y = x \]

2

Differentiate both sides implicitly:

\[ \frac{d}{dx}[e^y] = \frac{d}{dx}[x] \]

\[ e^y \cdot \frac{dy}{dx} = 1 \]

3

Solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{1}{e^y} \]

4

Substitute back using \( y = \ln(x) \):

\[ \frac{dy}{dx} = \frac{1}{e^{\ln(x)}} = \frac{1}{x} \]

Thus, \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \)

Derivatives of Common Inverse Functions

Function Inverse Function Derivative of Inverse
\( f(x) = x^2 \) (x ≥ 0) \( f^{-1}(x) = \sqrt{x} \) \( (f^{-1})'(x) = \frac{1}{2\sqrt{x}} \)
\( f(x) = e^x \) \( f^{-1}(x) = \ln(x) \) \( (f^{-1})'(x) = \frac{1}{x} \)
\( f(x) = \sin(x) \) (\( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \)) \( f^{-1}(x) = \arcsin(x) \) \( (f^{-1})'(x) = \frac{1}{\sqrt{1-x^2}} \)
\( f(x) = \cos(x) \) (\( 0 \leq x \leq \pi \)) \( f^{-1}(x) = \arccos(x) \) \( (f^{-1})'(x) = -\frac{1}{\sqrt{1-x^2}} \)
\( f(x) = \tan(x) \) (\( -\frac{\pi}{2} < x < \frac{\pi}{2} \)) \( f^{-1}(x) = \arctan(x) \) \( (f^{-1})'(x) = \frac{1}{1+x^2} \)

Inverse Function Practice

Test your understanding with these problems:

Common Mistakes to Avoid

Mistake Why It's Wrong Correct Approach
\( (f^{-1})'(x) = \frac{1}{f'(x)} \) Uses x instead of f⁻¹(x) in denominator \( (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \)
Forgetting domain restrictions Applies to functions that aren't one-to-one Always check if function is one-to-one on the domain
Confusing f⁻¹(x) with 1/f(x) Inverse function ≠ reciprocal Remember: f⁻¹ means functional inverse, not 1/f(x)
Not finding the correct corresponding point Uses wrong evaluation point Always find f⁻¹(x) first, then evaluate f' there

Quick Concept Check

Question 1: If \( f(2) = 5 \) and \( f'(2) = 3 \), what is \( (f^{-1})'(5) \)?

\( \frac{1}{3} \)
\( 3 \)
\( \frac{1}{5} \)
\( 5 \)
✓ Correct! \( (f^{-1})'(5) = \frac{1}{f'(f^{-1}(5))} = \frac{1}{f'(2)} = \frac{1}{3} \)
✗ Remember the formula: \( (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \)

Question 2: What is the derivative of \( \arcsin(x) \)?

\( \frac{1}{\sqrt{1-x^2}} \)
\( \frac{1}{1+x^2} \)
\( -\frac{1}{\sqrt{1-x^2}} \)
\( \sqrt{1-x^2} \)
✓ Correct! This is a standard inverse trig derivative.
✗ Remember: arcsin'(x) = 1/√(1-x²), arccos'(x) = -1/√(1-x²), arctan'(x) = 1/(1+x²)

Problem-Solving Strategy for Inverse Derivatives

  1. Identify the point on the inverse function
  2. Find the corresponding point on the original function
  3. Calculate the derivative of the original function
  4. Apply the inverse function theorem formula
  5. Verify your answer makes sense geometrically

AP Exam Tip

Inverse function derivatives frequently appear on AP Calculus exams. Remember to:

  • Memorize the derivatives of inverse trigonometric functions
  • Show the step of finding the corresponding point
  • Use the correct formula: \( (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \)
  • Check domain restrictions for inverse functions
  • Practice both the theorem and implicit differentiation methods

Practice Problems

Problem 1: Cubic Function Inverse

Let \( f(x) = x^3 + 1 \). Find \( (f^{-1})'(9) \).

Step 1: Find \( f^{-1}(9) \):

\( f(a) = 9 \Rightarrow a^3 + 1 = 9 \Rightarrow a^3 = 8 \Rightarrow a = 2 \)

So \( f^{-1}(9) = 2 \)

Step 2: Find \( f'(x) \):

\( f'(x) = 3x^2 \)

Step 3: Apply the theorem:

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

Problem 2: Exponential Inverse

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

Step 1: Find \( f^{-1}(1) \):

\( f(a) = 1 \Rightarrow e^{2a} = 1 \Rightarrow 2a = \ln(1) = 0 \Rightarrow a = 0 \)

So \( f^{-1}(1) = 0 \)

Step 2: Find \( f'(x) \):

\( f'(x) = 2e^{2x} \)

Step 3: Apply the theorem:

\( (f^{-1})'(1) = \frac{1}{f'(f^{-1}(1))} = \frac{1}{f'(0)} = \frac{1}{2e^{0}} = \frac{1}{2} \)

Problem 3: Using Implicit Differentiation

Use implicit differentiation to find the derivative of \( y = \arctan(x) \).

Step 1: Set up the inverse relationship:

\( y = \arctan(x) \Rightarrow \tan(y) = x \)

Step 2: Differentiate both sides:

\( \frac{d}{dx}[\tan(y)] = \frac{d}{dx}[x] \)

\( \sec^2(y) \cdot \frac{dy}{dx} = 1 \)

Step 3: Solve for \( \frac{dy}{dx} \):

\( \frac{dy}{dx} = \frac{1}{\sec^2(y)} = \cos^2(y) \)

Step 4: Express in terms of x:

Since \( \tan(y) = x \), we have a right triangle with opposite = x, adjacent = 1, hypotenuse = \( \sqrt{1+x^2} \)

\( \cos(y) = \frac{1}{\sqrt{1+x^2}} \Rightarrow \cos^2(y) = \frac{1}{1+x^2} \)

Thus, \( \frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2} \)

Back ⬅ Section 2.8 Next ➡ Section 2.10