Differentiating
Inverse Functions
Master the technique for finding derivatives of inverse functions using the Inverse Function Theorem and apply it to transcendental functions.
AP Calculus BC
What Is an Inverse Function?
If \( f \) is a one-to-one function, its inverse \( f^{-1} \) satisfies \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). Graphically, the curve of \( f^{-1} \) is the reflection of \( f \) across the line \( y = x \).
Reflection Across \( y = x \)
Why Differentiate Inverse Functions?
Implicit Relationships
Many functions are defined as inverses: \( \ln x \) is the inverse of \( e^x \), \( \arcsin x \) is the inverse of \( \sin x \).
Derivative Shortcuts
The Inverse Function Theorem lets you find \( (f^{-1})'(x) \) without ever solving for \( f^{-1} \) explicitly.
AP Exam Staple
This technique appears regularly on free-response and multiple-choice sections. It is a high-yield skill for the BC exam.
The Inverse Function Theorem
In words: the derivative of the inverse at a point is the reciprocal of the derivative of the original function, evaluated at the corresponding point.
Deriving the Formula
1
Start with the identity: \( f\!\left(f^{-1}(x)\right) = x \).
2
Differentiate both sides with respect to \( x \) using the chain rule: \( f'\!\left(f^{-1}(x)\right) \cdot \left(f^{-1}\right)'(x) = 1 \).
3
Solve for \( \left(f^{-1}\right)'(x) \): Divide both sides by \( f'\!\left(f^{-1}(x)\right) \) to obtain the theorem.
Geometric Interpretation
If \( f \) has slope \( m \) at a point, then \( f^{-1} \) has slope \( \tfrac{1}{m} \) at the reflected point. This is the geometric heart of the Inverse Function Theorem.
Key Distinction: \( (f^{-1})' \) vs. \( (f')^{-1} \)
| \( (f^{-1})'(x) \) | \( (f'(x))^{-1} \) | |
|---|---|---|
| Meaning | Derivative of the inverse function | Reciprocal of the derivative |
| Formula | \( \dfrac{1}{f'(f^{-1}(x))} \) | \( \dfrac{1}{f'(x)} \) |
| Evaluated at | The corresponding point on \( f \) | The same point \( x \) |
| Common error | Confusing these two: forgetting to compose with \( f^{-1}(x) \) | |
How to Apply the Theorem
1
Identify the target: You want \( (f^{-1})'(a) \) for some value \( a \).
2
Find the corresponding input: Determine \( b = f^{-1}(a) \), i.e., solve \( f(b) = a \).
3
Compute \( f'(b) \): Differentiate \( f \) and evaluate at \( b \).
4
Take the reciprocal: \( (f^{-1})'(a) = \dfrac{1}{f'(b)} \).
Example 1: Polynomial Inverse
Let \( f(x) = x^3 + x \). Find \( (f^{-1})'(2) \).
1
Solve \( f(b) = 2 \): Try \( b = 1 \): \( 1^3 + 1 = 2 \). ✓ So \( f^{-1}(2) = 1 \).
2
Compute \( f'(x) \): \( f'(x) = 3x^2 + 1 \).
3
Evaluate \( f'(1) \): \( f'(1) = 3(1)^2 + 1 = 4 \).
4
Reciprocal: \( (f^{-1})'(2) = \dfrac{1}{4} \).
Example 2: Table-Based Problem
Given the table of values, find \( (f^{-1})'(7) \).
| \( x \) | \( f(x) \) | \( f'(x) \) |
|---|---|---|
| 1 | 3 | 2 |
| 2 | 5 | 4 |
| 3 | 7 | 6 |
| 4 | 10 | 9 |
1
Find \( f^{-1}(7) \): From the table, \( f(3) = 7 \), so \( f^{-1}(7) = 3 \).
2
Look up \( f'(3) \): The table gives \( f'(3) = 6 \).
3
Result: \( (f^{-1})'(7) = \dfrac{1}{f'(3)} = \dfrac{1}{6} \).
Example 3: Deriving \( \frac{d}{dx}[\arcsin x] \) from the Theorem
Let \( f(x) = \sin x \) on \( \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] \), so \( f^{-1}(x) = \arcsin x \).
1
Apply the theorem: \( \frac{d}{dx}[\arcsin x] = \dfrac{1}{\cos(\arcsin x)} \).
2
Simplify \( \cos(\arcsin x) \): If \( \theta = \arcsin x \), then \( \sin\theta = x \). By the Pythagorean identity: \( \cos\theta = \sqrt{1 - x^2} \).
3
Final result: \( \dfrac{d}{dx}[\arcsin x] = \dfrac{1}{\sqrt{1-x^2}} \).
Inverse Trig Derivatives
| Function | Derivative |
|---|---|
| \( \arcsin x \) | \( \dfrac{1}{\sqrt{1-x^2}} \) |
| \( \arccos x \) | \( -\dfrac{1}{\sqrt{1-x^2}} \) |
| \( \arctan x \) | \( \dfrac{1}{1+x^2} \) |
| \( \text{arccot}\, x \) | \( -\dfrac{1}{1+x^2} \) |
| \( \text{arcsec}\, x \) | \( \dfrac{1}{|x|\sqrt{x^2-1}} \) |
| \( \text{arccsc}\, x \) | \( -\dfrac{1}{|x|\sqrt{x^2-1}} \) |
Deriving \( \frac{d}{dx}[\ln x] \) as an Inverse
Since \( \ln x \) is the inverse of \( e^x \), we let \( f(x) = e^x \) so that \( f^{-1}(x) = \ln x \).
1
Theorem gives: \( \frac{d}{dx}[\ln x] = \dfrac{1}{f'(f^{-1}(x))} = \dfrac{1}{e^{\ln x}} \).
2
Simplify: \( e^{\ln x} = x \), so \( \frac{d}{dx}[\ln x] = \dfrac{1}{x} \). ✓
Common Mistakes
✕ Wrong evaluation point
Writing \( (f^{-1})'(a) = \frac{1}{f'(a)} \) instead of \( \frac{1}{f'(f^{-1}(a))} \). You must first find \( f^{-1}(a) \).
✕ Forgetting restrictions
Inverse trig functions require restricted domains. Using the theorem outside the principal range yields incorrect signs or undefined results.
✕ Ignoring chain rule
When differentiating \( \arcsin(g(x)) \), students forget to multiply by \( g'(x) \). The inverse derivative formula alone is not enough.
Summary: Inverse Differentiation Toolkit
| Scenario | Strategy |
|---|---|
| Generic \( f^{-1} \) | \( (f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))} \) |
| Table of values | Look up \( f^{-1}(a) \), then read \( f' \) at that input |
| \( \arcsin x \), \( \arctan x \), etc. | Use memorized formulas (derived from the theorem) |
| \( \ln x \), \( \log_b x \) | Apply theorem to exponential inverse |
| Composition \( f^{-1}(g(x)) \) | Combine inverse derivative with chain rule |
Question 1 of 5
If \( f(x) = x^5 + 3x + 2 \), what is \( (f^{-1})'(6) \)?
Since \( f(1) = 1 + 3 + 2 = 6 \), we have \( f^{-1}(6) = 1 \). Then \( f'(x) = 5x^4 + 3 \), so \( f'(1) = 8 \). Thus \( (f^{-1})'(6) = \frac{1}{8} \).
Question 2 of 5
What is \( \frac{d}{dx}[\arctan(3x)] \) ?
By the chain rule: \( \frac{d}{dx}[\arctan(3x)] = \frac{1}{1+(3x)^2} \cdot 3 = \frac{3}{1+9x^2} \). Don't forget to multiply by the derivative of the inner function.
Question 3 of 5
Given the table below, find \( (g^{-1})'(4) \).
| \( x \) | \( g(x) \) | \( g'(x) \) |
|---|---|---|
| 1 | 4 | 3 |
| 2 | 8 | 5 |
| 4 | 12 | 7 |
Since \( g(1) = 4 \), we get \( g^{-1}(4) = 1 \). Then \( g'(1) = 3 \), so \( (g^{-1})'(4) = \frac{1}{g'(1)} = \frac{1}{3} \). A common mistake is using \( g'(4) = 7 \) instead.
Question 4 of 5
The derivative \( \frac{d}{dx}[\arcsin x] \) equals:
We showed earlier that if \( f(x) = \sin x \), then \( (f^{-1})'(x) = \frac{1}{\cos(\arcsin x)} = \frac{1}{\sqrt{1-x^2}} \). Choice (c) is the derivative of \( \arccos x \), not \( \arcsin x \).
Question 5 of 5
If \( h(x) = e^x + x \), what is \( (h^{-1})'(1) \)?
We need \( h(b) = 1 \). Try \( b = 0 \): \( e^0 + 0 = 1 \). So \( h^{-1}(1) = 0 \). Now \( h'(x) = e^x + 1 \), so \( h'(0) = 1 + 1 = 2 \). Therefore \( (h^{-1})'(1) = \frac{1}{2} \).