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.

more resources to help you study

practice multiple choice FRQ practice & scoring cheatsheets score calculator key terms

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

Forgetting the inner value. 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.
Confusing $(f^{-1})'(x)$ with $\frac{1}{f(x)}$ . The reciprocal applies to the derivative $f'$ , not to the function $f$ itself.
Assuming every function has a differentiable inverse. 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.
Mixing up which point to use. If $f(a) = b$ , the inverse passes through $(b, a)$ , not $(a, b)$ . The coordinates swap when you reflect across $y = x$ .
Thinking you always need the full inverse equation. You only need the single value $f^{-1}(x)$ at the point in question, not a formula for the whole inverse.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
chain rule	A differentiation rule that provides a method for finding the derivative of a composite function by multiplying the derivative of the outer function by the derivative of the inner function.
derivative	The instantaneous rate of change of a function at a specific point, representing the slope of the tangent line to the function at that point.
inverse function	A function that reverses the effect of another function, such that if f(a) = b, then the inverse function f⁻¹(b) = a.
inverse trigonometric functions	Functions that reverse the action of trigonometric functions, such as arcsine, arccosine, and arctangent, which return an angle given a trigonometric ratio.

Frequently Asked Questions

What is the derivative formula for an inverse function?

If f has a differentiable inverse, then the derivative of f inverse at x is 1 divided by f prime evaluated at f inverse of x. In words, evaluate the original derivative at the matching input, then take the reciprocal.

How do I find the matching input for an inverse derivative?

If f(a) = b, then f inverse of b equals a. For an inverse derivative at x = b, first find the original input a that produces b, then evaluate f prime at a.

Why are inverse function slopes reciprocals?

A function and its inverse reflect across the line y = x. At corresponding points, the tangent slopes swap rise and run, so the slopes are reciprocals when both derivatives exist.

When does the inverse derivative formula fail?

The formula fails when the original derivative at the matching input is zero, because division by zero is undefined. Geometrically, the inverse may have a vertical tangent there.

Do I need to solve for the whole inverse function?

Not always. Many AP Calculus problems only require the value of f inverse at one x-value, especially when information is given in a table.

How do inverse derivative questions show up on AP Calculus FRQs?

They often give tables of f, g, and their derivatives. Track the inverse value first, then use the reciprocal derivative formula and keep notation clear when writing tangent lines or slopes.