3.3 Differentiating Inverse Functions

To differentiate an inverse function, use $\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$. The slope of an inverse at a point is the reciprocal of the original function's slope at the matching point, because the two graphs are reflections across the line $y=x$. For AP Calculus, find the matching input before substituting into the formula.

Why This Matters for the AP Calculus Exam

Differentiating inverse functions builds directly on the chain rule and the definition of an inverse. This shows up on the AP Calculus exam in both multiple-choice and free-response questions, often using values pulled from tables or graphs instead of full equations. Many of these questions give you several functions at once (like $f$, $g$, and an inverse) and expect you to track which value goes where. Getting comfortable with the reciprocal relationship now also sets up Topic 3.4, where you derive the inverse trig derivatives.

Key Takeaways

• The derivative of an inverse is the reciprocal of the original derivative, evaluated at the matching input: $\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$.
• If $f(a) = b$, then $f^{-1}(b) = a$. Use this to find the inner value before plugging into the formula.
• The original function must be one-to-one (strictly increasing or decreasing) for the inverse to exist and be differentiable.
• The formula breaks down when $f'(f^{-1}(x)) = 0$, since you cannot divide by zero. This matches a vertical tangent on the inverse graph.
• Inverse function graphs are reflections across $y = x$, so their tangent slopes are reciprocals at corresponding points.

Differentiating Inverse Functions

If $f^{-1}(x)$ is the inverse of $f(x)$, a differentiable and invertible function, then the derivative of the inverse function can be found using this formula:

$\frac{d}{dx}\textcolor{blue}{f^{-1}(x)} = \frac{1}{\textcolor{red}{f'}(\textcolor{blue}{f^{-1}(x)})}$

This can also be written using $g(x)$ as the inverse of $f(x)$:

$\frac{d}{dx}g(x) = \frac{1}{f'(g(x))}$

A simple way to remember this rule: "the derivative of the inverse is the reciprocal of the derivative." This works because if $f(a) = b$, then $f^{-1}(b) = a$. If you want more practice with inverse functions first, review this Fiveable guide: Inverse Functions.

The slopes of a function and its inverse are reciprocals at matching points because the two graphs are reflections across the line $y = x$. At a point $(1, 8)$ on $f$, the reflected point $(8, 1)$ sits on $f^{-1}$, and their tangent slopes are reciprocals of each other.

Why the Formula Works

Start from the inverse relationship $f(f^{-1}(x)) = x$. Differentiate both sides with respect to $x$ using the chain rule:

$f'(f^{-1}(x)) \cdot \frac{d}{dx}f^{-1}(x) = 1$

Solving for the inverse derivative gives the formula. This is why the chain rule and the definition of an inverse are the two tools you need here.

How to Use This on the AP Calculus Exam

Problem Solving

When a question hands you $f(x)$ directly, you have two paths. You can solve for $f^{-1}(x)$ algebraically, or you can use the formula without ever finding the full inverse. The formula path is usually faster, especially when the inverse is messy to write out.

Example: If $f(x) = 2x + 1$, find $(f^{-1})'(1)$.

First find the inner value $f^{-1}(1)$. Switch variables in $y = 2x + 1$ to get $x = 2y + 1$, then solve:

$f^{-1}(x) = \frac{x-1}{2}$

So $f^{-1}(1) = \frac{1-1}{2} = 0$.

Now plug into the formula:

$\frac{d}{dx}f^{-1}(1) = \frac{1}{f'(f^{-1}(1))} = \frac{1}{f'(0)}$

Since $f'(x) = 2$, we get $f'(0) = 2$, so:

$\frac{d}{dx}f^{-1}(1) = \frac{1}{2}$

Free Response

Many free-response problems give you a table of values instead of equations. The skill is matching the right inputs.

Example (from the 2007 AP Calculus AB exam, College Board): The functions $f$ and $g$ are differentiable for all real numbers, and $g$ is strictly increasing. A table gives values of the functions and their first derivatives at selected $x$ values. If $g^{-1}$ is the inverse of $g$, write an equation for the line tangent to $y = g^{-1}(x)$ at $x = 2$.

Step 1: Find the inner value. You need $g^{-1}(2)$, which is not listed directly. Look for where $g(x) = 2$. Since $g(1) = 2$, you know $g^{-1}(2) = 1$.

Step 2: Apply the formula for the slope.

$\frac{d}{dx}g^{-1}(2) = \frac{1}{g'(g^{-1}(2))} = \frac{1}{g'(1)}$

With $g'(1) = 5$ from the table:

$\frac{d}{dx}g^{-1}(2) = \frac{1}{5}$

Step 3: Write the tangent line. Use point-slope form $y - y_1 = m(x - x_1)$. The slope is $m = \frac{1}{5}$, the point is $(2, 1)$ since $g^{-1}(2) = 1$:

$y - 1 = \frac{1}{5}(x - 2)$

This is a strong example of tracking three related functions at once: $g$, $g^{-1}$, and $g'$. Keeping your notation clear is important for showing your reasoning on free-response work.

Common Trap

The most common slip is plugging the original $x$ into $f'$ instead of plugging in $f^{-1}(x)$ first. The formula needs the derivative evaluated at the inner value, not at the outer input.

Common Misconceptions

• Misconception: The formula is $\frac{1}{f'(x)}$.
  Actually: The formula is $\frac{1}{f'(f^{-1}(x))}$, not $\frac{1}{f'(x)}$. You must find $f^{-1}(x)$ first, then evaluate the original derivative there.
• Misconception: $(f^{-1})'(x)$ is $\frac{1}{f(x)}$.
  Actually: The reciprocal applies to the derivative $f'$, not to the function $f$ itself.
• Misconception: Every function has a differentiable inverse.
  Actually: The original function must be one-to-one, and $f'$ must be nonzero at the matching point. If $f'(f^{-1}(x)) = 0$, the inverse has a vertical tangent and is not differentiable there.
• Misconception: If $f(a) = b$, the inverse passes through $(a, b)$.
  Actually: If $f(a) = b$, the inverse passes through $(b, a)$, not $(a, b)$. The coordinates swap when you reflect across $y = x$.
• Misconception: You always need the full inverse equation.
  Actually: You only need the single value $f^{-1}(x)$ at the point in question, not a formula for the whole inverse.

Related AP Calculus Guides

• 3.2 Implicit Differentiation
• 3.1 The Chain Rule
• 3.4 Differentiating Inverse Trigonometric Functions
• 3.6 Calculating Higher-Order Derivatives
• 3.5 Selecting Procedures for Calculating Derivatives
• Unit 3 Overview: Differentiation: Composite, Implicit, and Inverse Functions