Differentiating inverse trigonometric functions — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Derivations via implicit differentiation for derivatives of arcsine, arccosine, arctangent, arccotangent, arcsecant, arccosecant; application of the chain rule to composite inverse trigonometric functions; and simplification of derivatives using domain restrictions.

You should already know: Implicit differentiation for inverse functions. The chain rule for composite functions. Domain and range of principal-branch inverse trigonometric functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Differentiating inverse trigonometric functions?

This topic focuses on finding the instantaneous rate of change of inverse trigonometric functions, which return an angle (in radians) given a trigonometric ratio. Unlike regular trigonometric functions, inverse trig functions are defined with restricted domains to ensure they are one-to-one and invertible, and these restrictions directly impact the form and sign of their derivative formulas. According to the AP Calculus BC CED, this topic falls within Unit 3, which accounts for 9–13% of the total AP exam score. Questions on this topic appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with the chain rule, implicit differentiation, related rates, or integration problems. Mastery of this topic is required for later integration of inverse trigonometric functions, a high-weight topic in Unit 6. Unlike generic differentiation of basic polynomials or exponentials, differentiating inverse trig functions requires remembering domain-specific rules and correctly applying the chain rule when the argument of the inverse trig function is itself a function of $x$.

2. Deriving Derivatives of Basic Inverse Trigonometric Functions

The derivative formulas for inverse trigonometric functions can be derived directly from the definition of inverse functions and implicit differentiation, a skill you already mastered earlier in this unit. This process works identically for all six inverse trig functions: start by rewriting the inverse trig equation as a regular trig equation using the inverse function definition, differentiate both sides implicitly with respect to $x$, solve for $\frac{dy}{dx}$, then use a Pythagorean identity to rewrite $\frac{dy}{dx}$ in terms of $x$, adjusting the sign of the result to match the principal range of the inverse function.

For example, if $y=\arcsin x$, by definition $\sin y=x$ for $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$. Differentiating implicitly gives $\cos y\cdot \frac{dy}{dx}=1$, so $\frac{dy}{dx}=\frac{1}{\cos y}$. Using ${\sin }^{2}y+{\cos }^{2}y=1$, we get $\cos y=\sqrt{1-{\sin }^{2}y}=\sqrt{1-x^2}$. We take the positive root because $\cos y$ is positive for all $y$ in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, giving the final result: $$\frac{d}{dx}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}$$ The same process generates all six standard derivative formulas, with sign adjustments based on the principal range of each inverse function.

Worked Example

Problem: Derive the derivative of $y=\arctan x$ using implicit differentiation.

1. By definition, $y=\arctan x$ means $\tan y=x$, with principal range $-\frac{\pi }{2}<y<\frac{\pi }{2}$.
2. Differentiate both sides implicitly with respect to $x$: $$\frac{d}{dx}(\tan y)=\frac{d}{dx}(x)⟹{\sec }^{2}y\cdot \frac{dy}{dx}=1$$
3. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{1}{{\sec }^{2}y}$.
4. Use the Pythagorean identity $1+{\tan }^{2}y={\sec }^{2}y$, and substitute $\tan y=x$: $$\frac{dy}{dx}=\frac{1}{1+{\tan }^{2}y}=\frac{1}{1+x^2}$$
5. Confirm the sign: ${\sec }^{2}y$ is always positive for all $y$ in the principal range, so no sign adjustment is needed. The final derivative is $\frac{d}{dx}[\arctan x]=\frac{1}{1+x^2}$.

Exam tip: If you ever forget a derivative formula on the exam, you can re-derive it in 1–2 minutes using this implicit differentiation method, which is always accepted for full credit on FRQs.

3. Differentiating Composite Inverse Trigonometric Functions

Virtually all AP exam problems involving inverse trig derivatives require differentiating composite functions, where the argument of the inverse trig function is another function of $x$ (e.g., $\arcsin (3x^2)$, $\arctan (e^x)$, $\text{arcsec}(2x+1)$). For these functions, you must combine the inverse trig derivative rule with the chain rule: if $y=f(u(x))$ where $f$ is an inverse trig function, then $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$ This means you first apply the basic inverse trig derivative rule to $u(x)$, then multiply by the derivative of the inner function $u^{\prime }(x)$. A key requirement is that you substitute the entire inner function $u(x)$ into the inverse trig derivative formula (for example, for $y=\arctan (x^2)$, the denominator is $1+(x^2{)}^2=1+x^4$, not $1+x^2$).

Worked Example

Problem: Find $\frac{dy}{dx}$ for $y=\arccos (2x-1)$.

1. Identify the outer and inner functions: $y=\arccos (u)$, $u(x)=2x-1$, so $u^{\prime }(x)=2$.
2. Recall the derivative rule for $\arccos (u)$: $\frac{d}{dx}[\arccos (u)]=-\frac{u^{\prime }}{\sqrt{1-u^2}}$.
3. Substitute $u=2x-1$ and simplify the denominator: $$1-u^2=1-(2x-1{)}^2=1-(4x^2-4x+1)=4x-4x^2=4x(1-x)$$
4. Substitute back and simplify the radical: $$\frac{dy}{dx}=-\frac{2}{\sqrt{4x(1-x)}}=-\frac{2}{2\sqrt{x(1-x)}}=-\frac{1}{\sqrt{x(1-x)}}$$ This result is valid for $0<x<1$, the domain of the original function.

Exam tip: Always check that your final derivative is defined on the same domain as the original function, and simplify any perfect square factors from radicals—AP exam graders expect fully simplified expressions for full credit.

4. Differentiating Combinations of Inverse Trigonometric Functions

AP problems often require differentiating combinations of inverse trig functions with other types of functions (polynomials, exponentials, logarithms) using the product rule or quotient rule, just as you do for any other combination of functions. The process is identical to differentiating any other combination: first apply the product or quotient rule to set up your derivative expression, then compute the derivative of each inverse trig term using the inverse trig rule and chain rule. This tests your ability to combine multiple differentiation rules correctly, a core skill for the exam.

Worked Example

Problem: Find the slope of the tangent line to $f(x)=e^x\arcsin (x)$ at $x=0$.

1. $f(x)$ is the product of $e^x$ and $\arcsin (x)$, so apply the product rule: $$f^{\prime }(x)=\frac{d}{dx}(e^x)\cdot \arcsin (x)+e^x\cdot \frac{d}{dx}(\arcsin (x))$$
2. Substitute the derivatives of each term: $\frac{d}{dx}(e^x)=e^x$, $\frac{d}{dx}(\arcsin (x))=\frac{1}{\sqrt{1-x^2}}$, so: $$f^{\prime }(x)=e^x\arcsin (x)+\frac{e^x}{\sqrt{1-x^2}}$$
3. Evaluate $f^{\prime }(x)$ at $x=0$: $\arcsin (0)=0$, $e^0=1$, $\sqrt{1-0^2}=1$.
4. Simplify: $f^{\prime }(0)=(1\cdot 0)+\frac{1}{1}=1$. The slope of the tangent line at $x=0$ is 1.

Exam tip: When asked for the derivative at a point, always find the general derivative first, then substitute the point. Substituting the point early almost always leads to unnecessary errors.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Differentiating $y=\arcsin (5x)$ and writing $\frac{dy}{dx}=\frac{1}{\sqrt{1-(5x{)}^2}}$, omitting the chain rule term. Why: Students remember the basic derivative of $\arcsin x$ but forget the inner function $5x$ has a non-zero derivative, assuming $u^{\prime }=1$. Correct move: Always explicitly label the inner function $u(x)$ and write down $u^{\prime }$ before applying the inverse trig derivative rule, so you don't forget to include it as a factor.
• Wrong move: Writing the derivative of $\arccos (x)$ as $\frac{1}{\sqrt{1-x^2}}$, missing the negative sign. Why: Students mix up the derivatives of $\arcsin x$ and $\arccos x$, since they only differ by a sign. Correct move: Remember that $\arccos x$ is a decreasing function on its entire domain, so its derivative must always be negative—if you get a positive derivative for $\arccos$, you know you made a sign error.
• Wrong move: Differentiating $y=\text{arcsec}(3x)$ and writing $\frac{dy}{dx}=\frac{1}{3x\sqrt{9x^2-1}}$, omitting the absolute value around $3x$. Why: Students forget the derivative of $\text{arcsec}$ requires an absolute value from the domain restriction. Correct move: Always add an absolute value around the $u$ term when writing the derivative of $\text{arcsec}$ or $\text{arccsc}$, per the standard formula.
• Wrong move: Differentiating $y=\arctan (x^2)$ and writing $\frac{dy}{dx}=\frac{2x}{1+x^2}$, substituting $x$ instead of $x^2$ into the denominator. Why: Students rush and only substitute the inner function into the chain rule term, not into the inverse trig derivative formula. Correct move: After identifying $u(x)$, substitute $u$ into the inverse trig derivative formula before multiplying by $u^{\prime }$.
• Wrong move: Claiming $\frac{d}{dx}[\arcsin (\sin 2x)]=2$ for all real $x$. Why: Students assume $\arcsin (\sin f(x))=f(x)$ for all $x$, which is only true when $f(x)$ falls in the principal range of $\arcsin$. Correct move: Check the domain where the identity holds before differentiating, and only simplify if the problem specifies the interval of interest.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following is equal to $\frac{d}{dx}\left[\text{arcsec}(e^{2x})\right]$? A) $\frac{1}{e^{2x}\sqrt{e^{4x}-1}}$ B) $\frac{2}{\sqrt{e^{4x}-1}}$ C) $\frac{2e^{2x}}{\sqrt{e^{4x}-1}}$ D) $\frac{2}{e^{2x}\sqrt{e^{4x}-1}}$

Worked Solution: We use the chain rule for $y=\text{arcsec}(u)$, where $u=e^{2x}$. The derivative formula for $\text{arcsec}(u)$ is $\frac{d}{dx}[\text{arcsec}(u)]=\frac{u^{\prime }}{|u|\sqrt{u^2-1}}$. First compute $u^{\prime }=\frac{d}{dx}(e^{2x})=2e^{2x}$. Since $e^{2x}$ is always positive for all real $x$, $|u|=|e^{2x}|=e^{2x}$. Substitute into the formula: $\frac{dy}{dx}=\frac{2e^{2x}}{e^{2x}\sqrt{(e^{2x}{)}^2-1}}=\frac{2}{\sqrt{e^{4x}-1}}$ after canceling the $e^{2x}$ terms. The correct answer is B.

Question 2 (Free Response)

Consider the function $f(x)=\arctan (x^2-2x)$. (a) Find $f^{\prime }(x)$, the derivative of $f(x)$. (b) Find all critical points of $f(x)$ on the interval $0\le x\le 3$. (c) Classify each critical point from (b) as a local maximum, local minimum, or neither.

Worked Solution: (a) Use the chain rule with $u=x^2-2x$, so $u^{\prime }=2x-2$. The derivative of $\arctan (u)$ is $\frac{u^{\prime }}{1+u^2}$, so: $$f^{\prime }(x)=\frac{2x-2}{1+(x^2-2x{)}^2}=\frac{2(x-1)}{1+(x(x-2){)}^2}$$ (b) Critical points occur where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined. The denominator $1+(x^2-2x{)}^2$ is always positive for all real $x$, so $f^{\prime }(x)$ is defined everywhere on $[0,3]$. Set the numerator equal to zero: $2(x-1)=0⟹x=1$. The only critical point on $[0,3]$ is $x=1$. (c) Use the first derivative test: For $x<1$ (e.g., $x=0$), $f^{\prime }(0)=\frac{2(-1)}{1+0}=-2<0$. For $x>1$ (e.g., $x=2$), $f^{\prime }(2)=\frac{2(1)}{1+0}=2>0$. $f(x)$ changes from decreasing to increasing at $x=1$, so $f(x)$ has a local minimum at $x=1$.

Question 3 (Application / Real-World Style)

A camera is positioned 100 meters horizontally from the launch pad of a rocket that is rising straight up at a constant speed of 50 m/s. Let $\theta$ be the angle (in radians) between the camera's line of sight and the horizontal. Find the rate of change of $\theta$ (in radians per second) when the rocket is 100 meters above the ground.

Worked Solution: Let $h$ be the height of the rocket, so $\frac{dh}{dt}=50$ m/s, and the horizontal distance is constant at 100 m. By trigonometry, $\tan \theta =\frac{h}{100}$, so $\theta =\arctan \left(\frac{h}{100}\right)$. Differentiate both sides with respect to time $t$: $$\frac{d\theta }{dt}=\frac{1}{1+{\left(\frac{h}{100}\right)}^2}\cdot \frac{1}{100}\cdot \frac{dh}{dt}$$ Substitute $h=100$ m and $\frac{dh}{dt}=50$ m/s: $$\frac{d\theta }{dt}=\frac{1}{1+(1{)}^2}\cdot \frac{1}{100}\cdot 50=\frac{1}{2}\cdot \frac{50}{100}=\frac{1}{4}=0.25$$ When the rocket is 100 meters above the ground, the angle of the camera's line of sight is increasing at a rate of 0.25 radians per second.

7. Quick Reference Cheatsheet

8. What's Next

Differentiating inverse trigonometric functions is a critical prerequisite for integration techniques you will learn in Unit 6, where inverse trig derivatives are used to integrate functions of the form $\frac{1}{a^2+x^2}$, $\frac{1}{\sqrt{a^2-x^2}}$, and $\frac{1}{x\sqrt{x^2-a^2}}$. Recognizing an integrand matches the derivative of an inverse trig function is the core skill for these integration problems, which are commonly tested on both MCQ and FRQ sections. This topic also appears regularly in related rates problems, where you often need to differentiate an inverse trig angle with respect to time. Without mastering derivative formulas and chain rule application for inverse trig functions, you will not be able to correctly solve these common AP exam problems.

• Implicit Differentiation
• Chain Rule for Composite Functions
• Related Rates
• Inverse Trigonometric Integration