Differentiating inverse functions

AP Calculus BC· AP Calculus BC CED — Differentiation: Composite, Implicit, and Inverse Functions· 14 min read

1. The General Inverse Derivative Formula★★☆☆☆⏱ 3 min

📘 Definition

Derivative of an Inverse Function

(f^{-1})'(a)

If $f$ is a differentiable, one-to-one function on an interval, then for any point $a$ where $f^{'} (f^{- 1} (a)) \neq = 0$ , the derivative of $f^{- 1}$ at $a$ is given by the formula above.

Example:

Used to find the derivative of an inverse at a point without finding the general inverse function.

🔬 Derivation

Goal:

Derive the general inverse derivative formula

Starting from:

Definition of inverse function: $f (f^{- 1} (x)) = x$ for all $x$ in the domain of $f^{- 1}$

1
Differentiate both sides with respect to $x$ , applying the chain rule to the left-hand side:
2
$f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1$
3
Rearrange to isolate $(f^{- 1})^{'} (x)$ to get the final formula.

Result:

The derivative of an inverse at $a$ is the reciprocal of the derivative of the original function at $f^{- 1} (a)$

📐 Worked Example

Given $f (x) = 2 x^{3} + 3 x + 1$ , which is strictly increasing (and therefore one-to-one) for all real $x$ , find $(f^{- 1})^{'} (6)$ .

1
First find $f^{- 1} (6)$ , which is the value of $x$ such that $f (x) = 6$ :
2
$2x^3 + 3x + 1 = 6 \implies 2x^3 + 3x - 5 = 0$
3
Testing small integer values gives $x = 1$ as a solution, so $f^{- 1} (6) = 1$ .
4
Compute the derivative of the original function:
5
$f'(x) = 6x^2 + 3$
6
Evaluate $f^{'}$ at $f^{- 1} (6) = 1$ :
7
$f'(1) = 6(1)^2 + 3 = 9$
8
Apply the inverse derivative formula to get the result:
9
$(f^{-1})'(6) = \frac{1}{f'(f^{-1}(6))} = \frac{1}{9}$

Exam tip:

When finding $f^{- 1} (a)$ for polynomials on the AP exam, always test small integer values ( $x = - 2, - 1, 0, 1, 2$ ) first: exam problems are constructed so this value is always a small integer, so you never need to solve complicated higher-degree equations.

2. Differentiating Inverses via Implicit Differentiation★★☆☆☆⏱ 3 min

When you need the general derivative of an inverse function (not just the derivative at a single point), implicit differentiation is the most straightforward method. It works even when you cannot write the inverse function explicitly in terms of $x$ . The method follows directly from the definition of an inverse: if $y = f^{- 1} (x)$ , then $x = f (y)$ , where $y$ is a function of $x$ .

📐 Worked Example

Use implicit differentiation to derive the derivative of $f^{- 1} (x) = ln x$ , the inverse of $f (y) = e^{y}$ .

1
Let $y = ln x$ . By definition of inverse, $x = e^{y}$ for $x > 0$ .
2
Differentiate both sides with respect to $x$ : the left-hand side derivative is $1$ , and the right-hand side uses the chain rule:
3
$\frac{d}{dx}[x] = 1 = \frac{d}{dx}[e^y] = e^y \cdot \frac{dy}{dx}$
4
Rearrange to solve for $\frac{d y}{d x}$ :
5
$\frac{dy}{dx} = \frac{1}{e^y}$
6
Substitute back $x = e^{y}$ to get the derivative in terms of $x$ :
7
$\frac{d}{dx} \ln x = \frac{1}{x}, \quad x>0$

Exam tip:

If you forget the derivative of a specific inverse function (like $arcsin x$ ) on exam day, you can always re-derive it quickly using this implicit method, eliminating memorization errors.

3. Derivatives of Inverse Trigonometric Functions★★★☆☆⏱ 3 min

Inverse trigonometric functions are inverses of trigonometric functions restricted to domains that make them one-to-one. All inverse trigonometric functions have algebraic derivatives, which makes them extremely useful for integration later in the course. When the argument of an inverse trigonometric function is a function of $x$ , you must apply the chain rule just like for any other composite function.

📐 Worked Example

Find $\frac{d}{d x} [arctan (x^{2} + 1)]$ .

1
Recall the standard chain rule formula for $arctan (u)$ :
2
$\frac{d}{dx}\arctan(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$
3
Identify the inner function $u = x^{2} + 1$ , so:
4
$\frac{du}{dx} = 2x$
5
Substitute into the formula and simplify the denominator:
6
$\frac{d}{dx}\arctan(x^2 + 1) = \frac{1}{1 + (x^2 + 1)^2} \cdot 2x = \frac{2x}{x^4 + 2x^2 + 2}$

4. AP-Style Worked Practice Problems★★★☆☆⏱ 5 min

📐 Worked Example

Let $f (x) = x^{5} + 2 x^{3} + 4 x - 7$ , which is strictly increasing and one-to-one for all real $x$ . What is the value of $(f^{- 1})^{'} (0)$ ?

A) $\frac{1}{15}$ B) $\frac{1}{9}$ C) $9$ D) $15$

1
Use the inverse derivative formula $(f^{- 1})^{'} (a) = \frac{1}{f ^{'} ( f ^{- 1} ( a ))}$ with $a = 0$ .
2
First find $f^{- 1} (0)$ by solving $f (x) = 0$ . Testing small integers gives $f (1) = 1 + 2 + 4 - 7 = 0$ , so $f^{- 1} (0) = 1$ .
3
Compute the derivative of $f$ :
4
$f'(x) = 5x^4 + 6x^2 + 4$
5
Evaluate $f^{'}$ at $x = 1$ :
6
$f'(1) = 5(1) + 6(1) + 4 = 15$
7
Apply the formula to get the result:
8
$(f^{-1})'(0) = \frac{1}{15}, \text{ so the correct answer is A}$

📐 Worked Example

Let $f (x) = cos x$ , restricted to $[0, π]$ to make it one-to-one, with inverse $f^{- 1} (x) = arccos x$ .

(a) Use implicit differentiation to derive the general formula for $\frac{d}{d x} arccos x$ . (b) Evaluate $\frac{d}{d x} arccos x$ at $x = \frac{1}{2}$ . (c) Find the slope of the tangent line to $y = arccos (3 x)$ at $x = \frac{1}{6}$ .

1
Part (a): Let $y = arccos x$ , so by definition $x = cos y$ , with $0 \leq y \leq π$ and $- 1 \leq x \leq 1$ .
2
Differentiate both sides with respect to $x$ :
3
$1 = -\sin y \cdot \frac{dy}{dx}$
4
Solve for $\frac{d y}{d x}$ :
5
$\frac{dy}{dx} = -\frac{1}{\sin y}$
6
Use the Pythagorean identity: $sin y = 1 - cos^{2} y = 1 - x^{2}$ (positive root because $sin y \geq 0$ for $0 \leq y \leq π$ ). Thus:
7
$\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, \quad -1 < x < 1$
8
Part (b): Substitute $x = \frac{1}{2}$ into the formula:
9
$\frac{d}{dx}\arccos x \bigg|_{x=1/2} = -\frac{1}{\sqrt{1 - (1/2)^2}} = -\frac{2\sqrt{3}}{3}$
10
Part (c): Apply the chain rule, then evaluate at $x = \frac{1}{6}$ :
11
$\frac{d}{dx}\arccos(3x) = -\frac{3}{\sqrt{1 - 9x^2}}$
12
$9\left(\frac{1}{6}\right)^2 = \frac{1}{4} \implies \text{slope} = -\frac{3}{\sqrt{3/4}} = -2\sqrt{3}$

📐 Worked Example

In optics, the angle of refraction $θ_{2}$ (radians) of light passing from air to water is related to the angle of incidence $θ_{1}$ (radians) by Snell's Law: $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ , where $n_{1} = 1.00$ (air) and $n_{2} = 1.33$ (water). For $0 \leq θ_{1}, θ_{2} \leq \frac{π}{2}$ , we can write $θ_{1} = arcsin (1.33 sin θ_{2})$ . Find $\frac{d θ _{1}}{d θ _{2}}$ when $θ_{2} = \frac{π}{6}$ , and interpret the result.

1
Apply the chain rule to get the general derivative:
2
$\frac{d\theta_1}{d\theta_2} = \frac{1.33 \cos \theta_2}{\sqrt{1 - (1.33 \sin \theta_2)^2}}$
3
Substitute $θ_{2} = \frac{π}{6}$ : $sin (\frac{π}{6}) = 0.5$ , $cos (\frac{π}{6}) \approx 0.8660$
4
Calculate intermediate values:
5
$1.33 \sin \theta_2 = 0.665, \quad 1 - (0.665)^2 \approx 0.5578, \quad 1.33 \cos \theta_2 \approx 1.1518$
6
Compute the final result:
7
$\frac{d\theta_1}{d\theta_2} \approx 1.54$
8
Interpretation: When the angle of refraction is 30° ( $\frac{π}{6}$ radians), a 1-radian increase in $θ_{2}$ corresponds to an approximate 1.54-radian increase in the angle of incidence $θ_{1}$ .

5. Common Pitfalls

Wrong move:

Evaluating $f^{'} (a)$ instead of $f^{'} (f^{- 1} (a))$ when calculating $(f^{- 1})^{'} (a)$

Why:

Students mix up which point to plug into the original function's derivative, confusing the input $a$ for the inverse with the input for $f^{'}$ .

Correct move:

First find $f^{- 1} (a)$ (the $x$ -value of the original function that gives output $a$ ), then plug that $x$ -value into $f^{'}$ , not $a$ .

Wrong move:

Forgetting the chain rule when differentiating composite inverse trigonometric functions, e.g., writing $\frac{d}{d x} arcsin (2 x) = \frac{1}{1 - ( 2 x ) ^{2}}$

Why:

Students remember the standard derivative of the basic inverse function but ignore that the argument is a function of $x$ , not just $x$ itself.

Correct move:

Always multiply by the derivative of the inner function whenever differentiating any composite function, including composite inverse functions.

Wrong move:

Ignoring domain restrictions when writing the derivative of an inverse trigonometric function, e.g., writing $\frac{d}{d x} arcsec (x) = \frac{1}{x x ^{2} - 1}$ for all $x \neq = \pm 1$

Why:

Students focus only on the derivative formula and forget that the original inverse function is only defined on a restricted domain, so its derivative only exists on that same domain.

Correct move:

Always check the domain of the original inverse function before writing the derivative, and state the domain explicitly if asked.

Wrong move:

Claiming $(f^{- 1})^{'} (a) = 0$ because $f^{'} (f^{- 1} (a)) = 0$

Why:

Students forget the non-zero requirement for the inverse derivative formula.

Correct move:

If $f^{'} (f^{- 1} (a)) = 0$ , state that the derivative of the inverse at $a$ is undefined (the inverse has a vertical tangent at that point).

Wrong move:

Forgetting the reciprocal and writing $(f^{- 1})^{'} (a) = f^{'} (f^{- 1} (a))$

Why:

The inverse relationship of the function leads students to incorrectly skip the reciprocal step for the slope.

Correct move:

Remember that reflecting over $y = x$ inverts the tangent slope, so the derivative of the inverse must be the reciprocal of $f^{'} (f^{- 1} (a))$ .

6. Quick Reference Cheatsheet

Category	Formula	Notes
General Inverse Derivative (at a point)	$(f^{- 1})^{'} (a) = \frac{1}{f ^{'} ( f ^{- 1} ( a ))}$	Applies when $f$ is differentiable one-to-one, $f^{'} (f^{- 1} (a)) \neq = 0$
Implicit Differentiation for Inverses	If $x = f (y)$ , then $\frac{d y}{d x} = \frac{1}{f ^{'} ( y )}$	Use to derive general derivative formulas
Derivative of natural log	$\frac{d}{d x} ln x = \frac{1}{x}$	Only defined for $x > 0$
Derivative of $arcsin x$	$\frac{d}{d x} arcsin x = \frac{1}{1 - x ^{2}}$	Defined for $- 1 < x < 1$
Derivative of $arccos x$	$\frac{d}{d x} arccos x = - \frac{1}{1 - x ^{2}}$	Defined for $- 1 < x < 1$
Derivative of $arctan x$	$\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}$	Defined for all real $x$
Derivative of $arcsec x$	$\frac{d}{d x} arcsec x = \frac{1}{∣ x ∣ x ^{2} - 1}$	Defined for $∣ x ∣ > 1$
Chain Rule for Composite Inverses	$\frac{d}{d x} arcsin (u (x)) = \frac{u ^{'} ( x )}{1 - u ( x ) ^{2}}$	Same pattern applies to all inverse functions

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Derivative of inverse at a point
2022 · FRQ
Derive derivative of arccos

What's Next

Differentiating inverse functions is a critical prerequisite for almost all integration topics later in AP Calculus BC. The derivatives of inverse trigonometric functions you learned here form the foundation for inverse trigonometric integration, a key method for integrating rational functions that is heavily tested on both multiple-choice and free-response sections of the AP exam. You will also use the inverse derivative relationship when working with parametric and polar functions later in the course, where swapping input and output roles is common. Mastery of this topic is essential to correctly set up and evaluate integrals on the AP exam.