Unit 3Derivatives of

\arcsin x

\arccos x

\arctan x

, etc.

Concept 4.189Built for May 2027 AP Exam

Derivatives of $\arcsin x$ , $\arccos x$ , $\arctan x$ , etc.

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The three inverse trigonometric derivatives come from one idea: an inverse slope is a reciprocal. Since the derivative of sine is cosine, the derivative of arcsine is the reciprocal of a cosine, and writing that cosine in terms of $x$ produces the square root. That construction explains both the formula and its domain, which stops at $x = \pm 1$ because the sine curve is flat at its peaks. Arccosine differs from arcsine by a single minus sign, and arctangent has no square root at all because the tangent curve is never flat. Every version you meet on the exam is one of these three with a chain factor attached.

On the AP Exam

What you’ll see: A derivative of $\arcsin(3x)$ or $\arctan(x^2)$ , or an integral that reverses one of these forms. On free response the arcsine form also appears as a slope question about an inverse function.
What you’ll do: Recall the three base forms, attach the inner rate from the chain rule, and keep the domain restriction visible when the question asks where the derivative exists.
Memory hook: Reciprocal of a root, except tangent. Two of the three carry a square root; the third does not.
Use it on the exam: For MCQs: The distractors usually drop the inner rate or move the minus sign. Check the coefficient in the numerator before anything else. For FRQs: These derivatives rarely appear alone. When they do, the point is usually for the chain factor, so write the inner rate on its own line.

Concept Review

Three Forms and Their Domains

Figure 6 — The three derivatives side by side with their domains. Arcsine and arccosine share the square root and differ by a sign; arctangent has neither.

$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1+x^2}$

The first two are defined only for $\lvert x\rvert < 1$ . At the endpoints the square root vanishes, the slope is infinite, and the derivative does not exist. The third is defined for every real $x$ , and its derivative is positive everywhere.

Three cards for arcsin, arccos and arctan with their derivatives and domains, plus notes on the minus sign and the chain rule version

Where the Square Root Comes From

Apply the inverse function formula to $f(x)=\sin x$ at a point where $f(a)=b$ . Then

$\left(\arcsin\right)'(b) = \frac{1}{\cos a}, \qquad b = \sin a$

The cosine must be rewritten in terms of $b$ . On the range of arcsine, $\cos a = \sqrt{1-\sin^2 a} = \sqrt{1-b^2}$ , and the positive root is correct because cosine is positive on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ . Substituting gives the arcsine formula. The same argument with the sign of cosine reversed gives arccosine.

The Chain Rule Version

Function	Derivative	Inner rate
$\arcsin(3x)$	$\frac{3}{\sqrt{1-9x^2}}$	$3$
$\arccos(x^2)$	$-\frac{2x}{\sqrt{1-x^4}}$	$2x$
$\arctan(2x)$	$\frac{2}{1+4x^2}$	$2$
$\arctan\left(e^x\right)$	$\frac{e^x}{1+e^{2x}}$	$e^x$

The argument of the square root is always $1$ minus the square of the inner function, and the numerator is always the inner rate. Both facts come from the same substitution.

Reading the Domain Restriction

For $\arcsin(3x)$ the requirement is $\lvert 3x\rvert <1$ , that is $\lvert x\rvert <\frac{1}{3}$ . The restriction follows the argument rather than the variable, which is why a chain rule version has a narrower domain than the base form. A question that asks where the derivative exists is asking for that inequality.

Common Misconception Check

Myth 1: "The derivative of $\arcsin x$ is $\frac{1}{\cos x}$ ." Truth: That is the intermediate step, not the answer. The cosine must be converted to a function of $x$ , which produces the square root.
Myth 2: "Arccosine has the same derivative as arcsine." Truth: They differ by a minus sign, and the sign is scored. Cosine decreases on its principal range while sine increases.
Myth 3: "Arctangent also needs a square root." Truth: It does not. The tangent curve is never flat, so its reciprocal slope never blows up and no root appears.

Topic-Specific Study Method

Three Shapes, One Construction

Borrowed idea: Every inverse trig derivative is the reciprocal of a slope you already know. The sine curve is flat at its peaks, so arcsine blows up at the ends of its domain. That is why the square root forms carry a restriction. The cosine version differs only by a sign. Tangent is never flat anywhere, so arctangent has no restriction at all. Memorise the three shapes and derive the missing constants in the moment. Then apply the chain rule to the argument exactly as in the previous section.

Given	Base form	With the chain factor
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$	—
$\arcsin(5x)$	$\frac{1}{\sqrt{1-u^2}}$	$\frac{5}{\sqrt{1-25x^2}}$
$\arccos(4x)$	$-\frac{1}{\sqrt{1-u^2}}$	$-\frac{4}{\sqrt{1-16x^2}}$
$\arctan(x^3)$	$\frac{1}{1+u^2}$	$\frac{3x^2}{1+x^6}$

① Differentiate all three base forms and, for each, state the interval on which the derivative exists. The interval is part of the answer on a domain question.

② Differentiate $\arcsin(x^2)$ and check that the result is positive for $x>0$ and negative for $x<0$ . The inner rate explains the sign.

③ Differentiate $\arctan x$ and then $\arctan(-x)$ , and check that the second derivative is the negative of the first. Arctangent is odd, so its derivative should be even.

Training & Error Diagnosis

Core Archetype (Simplest Form): Differentiate $y = \arcsin(4x)$ .

Standard Execution:

Base form. The derivative of arcsine is $\frac{1}{\sqrt{1-u^2}}$ .

Inner rate. With $u = 4x$ the rate is 4.

Combine. $y' = \frac{4}{\sqrt{1-16x^2}}$ .

State the domain. The derivative exists for $\lvert x\rvert <\frac{1}{4}$ .

Exam Variation (Context / Parameter Shift): A curve is given by $y = \arctan(3x)$ . Find the slope at $x = 1$ and write the tangent line there.

Solution: $y' = \frac{3}{1+9x^2}$ , so at $x=1$ the slope is $\frac{3}{10}$ . Since $y(1)=\arctan 3 \approx 1.249$ , the tangent line is $y - 1.249 = 0.3(x-1)$ .

Common Traps & Failure Signals:

The missing numerator (Signal: a derivative with 1 on top where the inner rate is not 1): the chain factor was dropped.
The wrong square (Signal: $1-x^2$ under the root when the argument is $3x$ ): the inner function was not substituted into the base form.
The wandering minus (Signal: a minus on arcsine or on arctangent): the sign was copied from the wrong row of the table.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Sign Copied: You wrote the minus on arcsine instead of arccosine.
Argument Ignored: You left $x^2$ in the denominator when the argument was $3x$ , which produces a derivative of the wrong shape.
Domain Dropped: You reported a derivative at an input where the square root is not real, which is the one error on this page that produces a number for a quantity that genuinely does not exist.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; These three derivatives appear in multiple-choice items and inside supplied models, and no released BC free-response part from 2017 to 2026 scores one as a step
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; In 2025 Q1 the model contains an arctangent, but the derivative is supplied in the stem rather than derived
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Base Form

What is $\frac{d}{dx}\arcsin x$ ?

MCQ 2

With a Chain Factor

What is $\frac{d}{dx}\arctan(5x)$ ?

MCQ 3

The Sign and the Domain

Which statement about $f(x)=\arccos(2x)$ is correct?

FRQ 1 • A Slope and a Tangent

The curve $C$ is given by $y = \arcsin\left(\frac{x}{2}\right)$ for $-2 < x < 2$ .

(a) Find $\frac{dy}{dx}$ .

(b) Find the slope of $C$ at $x = 1$ .

(c) Write an equation for the tangent line to $C$ at $x = 1$ .

(d) Explain why the derivative does not exist at $x = 2$ , even though the function itself is defined there.

Cheatsheet

Inverse Trigonometric Derivatives

	Derivative	Domain
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$	$\lvert x\rvert <1$
$\arccos x$	$-\frac{1}{\sqrt{1-x^2}}$	$\lvert x\rvert <1$
$\arctan x$	$\frac{1}{1+x^2}$	all real $x$
$\arcsin u$	$\frac{u'}{\sqrt{1-u^2}}$	$\lvert u\rvert <1$
$\arctan u$	$\frac{u'}{1+u^2}$	all real $u$

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Start Topic Practice Next Topic: 3.7 Combination of Rules

Unit 3Derivatives of

\arcsin x

\arccos x

\arctan x

, etc.

Concept 4.189Built for May 2027 AP Exam

Derivatives of $\arcsin x$ , $\arccos x$ , $\arctan x$ , etc.

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A derivative of $\arcsin(3x)$ or $\arctan(x^2)$ , or an integral that reverses one of these forms. On free response the arcsine form also appears as a slope question about an inverse function.
What you’ll do: Recall the three base forms, attach the inner rate from the chain rule, and keep the domain restriction visible when the question asks where the derivative exists.
Memory hook: Reciprocal of a root, except tangent. Two of the three carry a square root; the third does not.
Use it on the exam: For MCQs: The distractors usually drop the inner rate or move the minus sign. Check the coefficient in the numerator before anything else. For FRQs: These derivatives rarely appear alone. When they do, the point is usually for the chain factor, so write the inner rate on its own line.

Concept Review

Three Forms and Their Domains

Figure 6 — The three derivatives side by side with their domains. Arcsine and arccosine share the square root and differ by a sign; arctangent has neither.

$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1+x^2}$

Three cards for arcsin, arccos and arctan with their derivatives and domains, plus notes on the minus sign and the chain rule version

Where the Square Root Comes From

Apply the inverse function formula to $f(x)=\sin x$ at a point where $f(a)=b$ . Then

$\left(\arcsin\right)'(b) = \frac{1}{\cos a}, \qquad b = \sin a$

The Chain Rule Version

Function	Derivative	Inner rate
$\arcsin(3x)$	$\frac{3}{\sqrt{1-9x^2}}$	$3$
$\arccos(x^2)$	$-\frac{2x}{\sqrt{1-x^4}}$	$2x$
$\arctan(2x)$	$\frac{2}{1+4x^2}$	$2$
$\arctan\left(e^x\right)$	$\frac{e^x}{1+e^{2x}}$	$e^x$

The argument of the square root is always $1$ minus the square of the inner function, and the numerator is always the inner rate. Both facts come from the same substitution.

Reading the Domain Restriction

Common Misconception Check

Myth 1: "The derivative of $\arcsin x$ is $\frac{1}{\cos x}$ ." Truth: That is the intermediate step, not the answer. The cosine must be converted to a function of $x$ , which produces the square root.
Myth 2: "Arccosine has the same derivative as arcsine." Truth: They differ by a minus sign, and the sign is scored. Cosine decreases on its principal range while sine increases.
Myth 3: "Arctangent also needs a square root." Truth: It does not. The tangent curve is never flat, so its reciprocal slope never blows up and no root appears.

Topic-Specific Study Method

Three Shapes, One Construction

Given	Base form	With the chain factor
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$	—
$\arcsin(5x)$	$\frac{1}{\sqrt{1-u^2}}$	$\frac{5}{\sqrt{1-25x^2}}$
$\arccos(4x)$	$-\frac{1}{\sqrt{1-u^2}}$	$-\frac{4}{\sqrt{1-16x^2}}$
$\arctan(x^3)$	$\frac{1}{1+u^2}$	$\frac{3x^2}{1+x^6}$

① Differentiate all three base forms and, for each, state the interval on which the derivative exists. The interval is part of the answer on a domain question.

② Differentiate $\arcsin(x^2)$ and check that the result is positive for $x>0$ and negative for $x<0$ . The inner rate explains the sign.

③ Differentiate $\arctan x$ and then $\arctan(-x)$ , and check that the second derivative is the negative of the first. Arctangent is odd, so its derivative should be even.

Training & Error Diagnosis

Core Archetype (Simplest Form): Differentiate $y = \arcsin(4x)$ .

Standard Execution:

Base form. The derivative of arcsine is $\frac{1}{\sqrt{1-u^2}}$ .

Inner rate. With $u = 4x$ the rate is 4.

Combine. $y' = \frac{4}{\sqrt{1-16x^2}}$ .

State the domain. The derivative exists for $\lvert x\rvert <\frac{1}{4}$ .

Exam Variation (Context / Parameter Shift): A curve is given by $y = \arctan(3x)$ . Find the slope at $x = 1$ and write the tangent line there.

Solution: $y' = \frac{3}{1+9x^2}$ , so at $x=1$ the slope is $\frac{3}{10}$ . Since $y(1)=\arctan 3 \approx 1.249$ , the tangent line is $y - 1.249 = 0.3(x-1)$ .

Common Traps & Failure Signals:

The missing numerator (Signal: a derivative with 1 on top where the inner rate is not 1): the chain factor was dropped.
The wrong square (Signal: $1-x^2$ under the root when the argument is $3x$ ): the inner function was not substituted into the base form.
The wandering minus (Signal: a minus on arcsine or on arctangent): the sign was copied from the wrong row of the table.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Sign Copied: You wrote the minus on arcsine instead of arccosine.
Argument Ignored: You left $x^2$ in the denominator when the argument was $3x$ , which produces a derivative of the wrong shape.
Domain Dropped: You reported a derivative at an input where the square root is not real, which is the one error on this page that produces a number for a quantity that genuinely does not exist.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; These three derivatives appear in multiple-choice items and inside supplied models, and no released BC free-response part from 2017 to 2026 scores one as a step
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; In 2025 Q1 the model contains an arctangent, but the derivative is supplied in the stem rather than derived
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Base Form

What is $\frac{d}{dx}\arcsin x$ ?

MCQ 2

With a Chain Factor

What is $\frac{d}{dx}\arctan(5x)$ ?

MCQ 3

The Sign and the Domain

Which statement about $f(x)=\arccos(2x)$ is correct?

FRQ 1 • A Slope and a Tangent

The curve $C$ is given by $y = \arcsin\left(\frac{x}{2}\right)$ for $-2 < x < 2$ .

(a) Find $\frac{dy}{dx}$ .

(b) Find the slope of $C$ at $x = 1$ .

(c) Write an equation for the tangent line to $C$ at $x = 1$ .

(d) Explain why the derivative does not exist at $x = 2$ , even though the function itself is defined there.

Cheatsheet

Inverse Trigonometric Derivatives

	Derivative	Domain
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$	$\lvert x\rvert <1$
$\arccos x$	$-\frac{1}{\sqrt{1-x^2}}$	$\lvert x\rvert <1$
$\arctan x$	$\frac{1}{1+x^2}$	all real $x$
$\arcsin u$	$\frac{u'}{\sqrt{1-u^2}}$	$\lvert u\rvert <1$
$\arctan u$	$\frac{u'}{1+u^2}$	all real $u$

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Start Topic Practice Next Topic: 3.7 Combination of Rules

Unit 3Derivatives of

\arcsin x

\arccos x

\arctan x

, etc.

Concept 4.189Built for May 2027 AP Exam

Derivatives of $\arcsin x$ , $\arccos x$ , $\arctan x$ , etc.

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A derivative of $\arcsin(3x)$ or $\arctan(x^2)$ , or an integral that reverses one of these forms. On free response the arcsine form also appears as a slope question about an inverse function.
What you’ll do: Recall the three base forms, attach the inner rate from the chain rule, and keep the domain restriction visible when the question asks where the derivative exists.
Memory hook: Reciprocal of a root, except tangent. Two of the three carry a square root; the third does not.
Use it on the exam: For MCQs: The distractors usually drop the inner rate or move the minus sign. Check the coefficient in the numerator before anything else. For FRQs: These derivatives rarely appear alone. When they do, the point is usually for the chain factor, so write the inner rate on its own line.

Concept Review

Three Forms and Their Domains

Figure 6 — The three derivatives side by side with their domains. Arcsine and arccosine share the square root and differ by a sign; arctangent has neither.

$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1+x^2}$

Three cards for arcsin, arccos and arctan with their derivatives and domains, plus notes on the minus sign and the chain rule version

Where the Square Root Comes From

Apply the inverse function formula to $f(x)=\sin x$ at a point where $f(a)=b$ . Then

$\left(\arcsin\right)'(b) = \frac{1}{\cos a}, \qquad b = \sin a$

The Chain Rule Version

Function	Derivative	Inner rate
$\arcsin(3x)$	$\frac{3}{\sqrt{1-9x^2}}$	$3$
$\arccos(x^2)$	$-\frac{2x}{\sqrt{1-x^4}}$	$2x$
$\arctan(2x)$	$\frac{2}{1+4x^2}$	$2$
$\arctan\left(e^x\right)$	$\frac{e^x}{1+e^{2x}}$	$e^x$

The argument of the square root is always $1$ minus the square of the inner function, and the numerator is always the inner rate. Both facts come from the same substitution.

Reading the Domain Restriction

Common Misconception Check

Myth 1: "The derivative of $\arcsin x$ is $\frac{1}{\cos x}$ ." Truth: That is the intermediate step, not the answer. The cosine must be converted to a function of $x$ , which produces the square root.
Myth 2: "Arccosine has the same derivative as arcsine." Truth: They differ by a minus sign, and the sign is scored. Cosine decreases on its principal range while sine increases.
Myth 3: "Arctangent also needs a square root." Truth: It does not. The tangent curve is never flat, so its reciprocal slope never blows up and no root appears.

Topic-Specific Study Method

Three Shapes, One Construction

Given	Base form	With the chain factor
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$	—
$\arcsin(5x)$	$\frac{1}{\sqrt{1-u^2}}$	$\frac{5}{\sqrt{1-25x^2}}$
$\arccos(4x)$	$-\frac{1}{\sqrt{1-u^2}}$	$-\frac{4}{\sqrt{1-16x^2}}$
$\arctan(x^3)$	$\frac{1}{1+u^2}$	$\frac{3x^2}{1+x^6}$

① Differentiate all three base forms and, for each, state the interval on which the derivative exists. The interval is part of the answer on a domain question.

② Differentiate $\arcsin(x^2)$ and check that the result is positive for $x>0$ and negative for $x<0$ . The inner rate explains the sign.

③ Differentiate $\arctan x$ and then $\arctan(-x)$ , and check that the second derivative is the negative of the first. Arctangent is odd, so its derivative should be even.

Training & Error Diagnosis

Core Archetype (Simplest Form): Differentiate $y = \arcsin(4x)$ .

Standard Execution:

Base form. The derivative of arcsine is $\frac{1}{\sqrt{1-u^2}}$ .

Inner rate. With $u = 4x$ the rate is 4.

Combine. $y' = \frac{4}{\sqrt{1-16x^2}}$ .

State the domain. The derivative exists for $\lvert x\rvert <\frac{1}{4}$ .

Exam Variation (Context / Parameter Shift): A curve is given by $y = \arctan(3x)$ . Find the slope at $x = 1$ and write the tangent line there.

Solution: $y' = \frac{3}{1+9x^2}$ , so at $x=1$ the slope is $\frac{3}{10}$ . Since $y(1)=\arctan 3 \approx 1.249$ , the tangent line is $y - 1.249 = 0.3(x-1)$ .

Common Traps & Failure Signals:

The missing numerator (Signal: a derivative with 1 on top where the inner rate is not 1): the chain factor was dropped.
The wrong square (Signal: $1-x^2$ under the root when the argument is $3x$ ): the inner function was not substituted into the base form.
The wandering minus (Signal: a minus on arcsine or on arctangent): the sign was copied from the wrong row of the table.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Sign Copied: You wrote the minus on arcsine instead of arccosine.
Argument Ignored: You left $x^2$ in the denominator when the argument was $3x$ , which produces a derivative of the wrong shape.
Domain Dropped: You reported a derivative at an input where the square root is not real, which is the one error on this page that produces a number for a quantity that genuinely does not exist.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; These three derivatives appear in multiple-choice items and inside supplied models, and no released BC free-response part from 2017 to 2026 scores one as a step
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; In 2025 Q1 the model contains an arctangent, but the derivative is supplied in the stem rather than derived
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Base Form

What is $\frac{d}{dx}\arcsin x$ ?

MCQ 2

With a Chain Factor

What is $\frac{d}{dx}\arctan(5x)$ ?

MCQ 3

The Sign and the Domain

Which statement about $f(x)=\arccos(2x)$ is correct?

FRQ 1 • A Slope and a Tangent

The curve $C$ is given by $y = \arcsin\left(\frac{x}{2}\right)$ for $-2 < x < 2$ .

(a) Find $\frac{dy}{dx}$ .

(b) Find the slope of $C$ at $x = 1$ .

(c) Write an equation for the tangent line to $C$ at $x = 1$ .

(d) Explain why the derivative does not exist at $x = 2$ , even though the function itself is defined there.

Cheatsheet

Inverse Trigonometric Derivatives

	Derivative	Domain
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$	$\lvert x\rvert <1$
$\arccos x$	$-\frac{1}{\sqrt{1-x^2}}$	$\lvert x\rvert <1$
$\arctan x$	$\frac{1}{1+x^2}$	all real $x$
$\arcsin u$	$\frac{u'}{\sqrt{1-u^2}}$	$\lvert u\rvert <1$
$\arctan u$	$\frac{u'}{1+u^2}$	all real $u$

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