Unit 3: Differentiation: Composite, Implicit, and Inverse FunctionsDerivatives of Inverse Trigonometric Functions

Concept 4.136Built for May 2027 AP Exam

Derivatives of Inverse Trigonometric Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

There are six inverse trigonometric derivatives, and they arrive in three pairs. Each pair differs only by a sign, which is why the formulas are easy to mix up under time pressure. The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$ , and the derivative of $\arccos x$ is its negative. Every one of these formulas is silent about its domain, and the square root or the absolute value in the denominator is the only reminder that a restriction exists. Read the restriction first. Then differentiate. A composite argument adds one ordinary chain factor on top of whichever formula you selected.

On the AP Exam

The three formulas are short, and the exam treats them that way. In nine released exams no free-response part differentiates an inverse trigonometric function: $\arctan$ turns up twice, once as a value sitting inside a limit and once inside a model whose derivative is handed to you. This is multiple-choice territory.

What you’ll see: $\arcsin$ , $\arccos$ , $\arctan$ , or $\operatorname{arcsec}$ applied to a bare $x$ or to a linear expression, usually inside a sum or a product.
What you’ll do: Select the correct formula from the pair, then multiply by the derivative of the argument.
Memory hook: Arcsin up, arccos down, arctan flat. The three shapes of the denominators follow that order.
Use it on the exam: For MCQs: The distractor with the sign of the partner function appears in every version, so confirm which of the pair you are differentiating before choosing. For FRQs: No released part scores these directly, but the reason to know them survives: when an inverse trig term sits inside a larger derivative it has to disappear when you differentiate it, and an answer still carrying one is unfinished.

Concept Review

Six Formulas, Three Pairs

Figure 1 — The six inverse trigonometric derivatives reduce to three denominator shapes, and each pair differs only by a sign. The domain restriction lives in the denominator, not in the formula's outline.

SIX INVERSE TRIG SLOPES, THREE PAIRS

The Table Worth Memorising

Function	Derivative	Domain of the derivative
$\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\arccos x$	$-\dfrac{1}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\arctan x$	$\dfrac{1}{1+x^2}$	All real $x$
$\operatorname{arccot} x$	$-\dfrac{1}{1+x^2}$	All real $x$
$\operatorname{arcsec} x$	$\dfrac{1}{\lvert x \rvert\sqrt{x^2-1}}$	$\lvert x \rvert > 1$
$\operatorname{arccsc} x$	$-\dfrac{1}{\lvert x \rvert\sqrt{x^2-1}}$	$\lvert x \rvert > 1$

The absolute value in the secant and cosecant rows is not decoration. Without it the formula gives the wrong sign on the negative branch, and the exam tests that branch deliberately.

Composing With a Chain Factor

For $\arcsin(3x)$ the formula supplies $\frac{1}{\sqrt{1-(3x)^2}}$ and the argument supplies a factor of $3$ , so the derivative is $\frac{3}{\sqrt{1-9x^2}}$ . The argument must be substituted into the formula completely before the chain factor is attached; writing $\sqrt{1-x^2}$ in the denominator is the single most common error in this topic.

Common Misconception Check

Myth 1: " $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}[\arccos x]$ is the same." Truth: The arccos derivative carries a minus sign, and that sign propagates through every later term.
Myth 2: "The chain factor goes inside the square root." Truth: The chain factor multiplies the whole fraction, and the argument is squared inside the root.
Myth 3: " $\operatorname{arcsec}$ follows the same pattern as $\arcsin$ ." Truth: Its denominator carries an absolute value and a $\sqrt{x^2-1}$ , and the two branches have to be handled separately.

Topic-Specific Study Method

Pick the Pair, Then Chain

Borrowed idea: There are six inverse trigonometric derivatives, and they arrive in three pairs. Each pair differs only by a sign, which is why the formulas are easy to mix up under time pressure. The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$ , and the derivative of $\arccos x$ is its negative. Every one of these formulas is silent about its domain, and the square root or the absolute value in the denominator is the only reminder that a restriction exists. Read the restriction first. Then differentiate.

Exam disguise	What you see on the page	What your brain must translate	Required execution
A bare argument	$\arcsin x$	One formula, no chain factor	Write the formula and stop
A linear argument	$\arctan 4x$	The formula plus a constant factor	Substitute $4x$ , multiply by $4$
A composite argument	$\arcsin(x^2)$	Substitute the whole argument first	Square $x^2$ , then chain $2x$
A sum with a trig term	$\arcsin x + \sqrt{1-x^2}$	Two independent terms	Differentiate each, keep the signs
A secant form	$\operatorname{arcsec} x$	Absolute value and two branches	Handle each branch, or state the restriction

① Write the six formulas from memory three times in a row, each time grouping them into the three pairs. The pairing is what makes the signs survive an exam.

② Differentiate $\arcsin(3x)$ , $\arctan(3x)$ , and $\operatorname{arcsec}(3x)$ side by side and compare where the chain factor lands in each.

③ On any composite inverse-trig term, substitute the argument into the formula completely before attaching anything else. The substitution is the step that gets rushed, and the error it produces is invisible afterwards.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find $\frac{d}{dx}\left[\arcsin(3x)\right]$ .

Standard Execution:

Select the formula. The function is arcsin, so the denominator is $\sqrt{1-u^2}$ .

Substitute the argument. With $u = 3x$ the denominator becomes $\sqrt{1-9x^2}$ .

Attach the chain factor. The derivative of $3x$ is $3$ , which multiplies the fraction.

Write the answer. $\frac{3}{\sqrt{1-9x^2}}$ .

State the restriction. The expression requires $|x| < \frac{1}{3}$ .

Exam Variation (Context / Parameter Shift): Let $f(x) = \arctan(x^2)$ . Find $f'(x)$ and $f'(1)$ .

Solution: With $u = x^2$ , the arctan formula gives $\frac{1}{1+x^4}$ , and the chain factor is $2x$ . So $f'(x) = \frac{2x}{1+x^4}$ and $f'(1) = \frac{2}{2} = 1$ . Note that the argument is squared inside the denominator and the chain factor stays outside the fraction; swapping those two placements is the standard error.

Common Traps & Failure Signals:

The unsubstituted argument (Signal: an $x^2$ where $9x^2$ belongs): the formula was written for the bare $x$ and never updated.
The sign of the partner (Signal: a positive arccos derivative): the pair was selected by shape rather than by name.
The swallowed chain factor (Signal: a factor of $1$ where the argument had a slope): the argument was substituted and its derivative was never multiplied in.

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

Pair Confusion $\rightarrow$ Say the function name before writing the formula.
Argument Shortcut: You copied the formula in its bare- $x$ form because that is the version memorised, and the substitution step felt redundant once the shape looked familiar.
Sign Drift: You kept the arctan sign while differentiating arccos.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The three formulas are judged in multiple choice; the two $\arctan$ appearances in the released sets are a limit value and a model whose derivative is supplied
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part; no scored part differentiates an inverse trigonometric function
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

A Linear Argument

What is the derivative of $f(x) = \arcsin(3x)$ ?

MCQ 2

A Composite Argument

Let $g(x) = \arctan(x^2)$ . What is $g'(x)$ ?

MCQ 3

The Sign of the Pair

What is the derivative of $h(x) = \arccos x$ ?

FRQ 4

An Inverse Tangent and a Tangent Line

Let $f(x) = \arctan(2x)$ .

(a) Find $f'(x)$ .

(b) Find the equation of the line tangent to the graph of $f$ at $x = \frac{1}{2}$ .

(c) Let $h(x) = f(x^2)$ . Find $h'(x)$ and evaluate $h'(1)$ .

(d) Explain why $f'(x)$ is positive for every real $x$ , and what that says about the graph of $f$ .

Cheatsheet

Inverse Trigonometric Derivatives

	Derivative
$\arcsin u$	$\dfrac{u'}{\sqrt{1-u^2}}$
$\arccos u$	$-\dfrac{u'}{\sqrt{1-u^2}}$
$\arctan u$	$\dfrac{u'}{1+u^2}$
$\operatorname{arccot} u$	$-\dfrac{u'}{1+u^2}$
$\operatorname{arcsec} u$	$\dfrac{u'}{\lvert u \rvert\sqrt{u^2-1}}$
$\operatorname{arccsc} u$	$-\dfrac{u'}{\lvert u \rvert\sqrt{u^2-1}}$

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Unit 3: Differentiation: Composite, Implicit, and Inverse FunctionsDerivatives of Inverse Trigonometric Functions

Concept 4.136Built for May 2027 AP Exam

Derivatives of Inverse Trigonometric Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: $\arcsin$ , $\arccos$ , $\arctan$ , or $\operatorname{arcsec}$ applied to a bare $x$ or to a linear expression, usually inside a sum or a product.
What you’ll do: Select the correct formula from the pair, then multiply by the derivative of the argument.
Memory hook: Arcsin up, arccos down, arctan flat. The three shapes of the denominators follow that order.
Use it on the exam: For MCQs: The distractor with the sign of the partner function appears in every version, so confirm which of the pair you are differentiating before choosing. For FRQs: No released part scores these directly, but the reason to know them survives: when an inverse trig term sits inside a larger derivative it has to disappear when you differentiate it, and an answer still carrying one is unfinished.

Concept Review

Six Formulas, Three Pairs

SIX INVERSE TRIG SLOPES, THREE PAIRS

The Table Worth Memorising

Function	Derivative	Domain of the derivative
$\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\arccos x$	$-\dfrac{1}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\arctan x$	$\dfrac{1}{1+x^2}$	All real $x$
$\operatorname{arccot} x$	$-\dfrac{1}{1+x^2}$	All real $x$
$\operatorname{arcsec} x$	$\dfrac{1}{\lvert x \rvert\sqrt{x^2-1}}$	$\lvert x \rvert > 1$
$\operatorname{arccsc} x$	$-\dfrac{1}{\lvert x \rvert\sqrt{x^2-1}}$	$\lvert x \rvert > 1$

The absolute value in the secant and cosecant rows is not decoration. Without it the formula gives the wrong sign on the negative branch, and the exam tests that branch deliberately.

Composing With a Chain Factor

Common Misconception Check

Myth 1: " $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}[\arccos x]$ is the same." Truth: The arccos derivative carries a minus sign, and that sign propagates through every later term.
Myth 2: "The chain factor goes inside the square root." Truth: The chain factor multiplies the whole fraction, and the argument is squared inside the root.
Myth 3: " $\operatorname{arcsec}$ follows the same pattern as $\arcsin$ ." Truth: Its denominator carries an absolute value and a $\sqrt{x^2-1}$ , and the two branches have to be handled separately.

Topic-Specific Study Method

Pick the Pair, Then Chain

Exam disguise	What you see on the page	What your brain must translate	Required execution
A bare argument	$\arcsin x$	One formula, no chain factor	Write the formula and stop
A linear argument	$\arctan 4x$	The formula plus a constant factor	Substitute $4x$ , multiply by $4$
A composite argument	$\arcsin(x^2)$	Substitute the whole argument first	Square $x^2$ , then chain $2x$
A sum with a trig term	$\arcsin x + \sqrt{1-x^2}$	Two independent terms	Differentiate each, keep the signs
A secant form	$\operatorname{arcsec} x$	Absolute value and two branches	Handle each branch, or state the restriction

① Write the six formulas from memory three times in a row, each time grouping them into the three pairs. The pairing is what makes the signs survive an exam.

② Differentiate $\arcsin(3x)$ , $\arctan(3x)$ , and $\operatorname{arcsec}(3x)$ side by side and compare where the chain factor lands in each.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find $\frac{d}{dx}\left[\arcsin(3x)\right]$ .

Standard Execution:

Select the formula. The function is arcsin, so the denominator is $\sqrt{1-u^2}$ .

Substitute the argument. With $u = 3x$ the denominator becomes $\sqrt{1-9x^2}$ .

Attach the chain factor. The derivative of $3x$ is $3$ , which multiplies the fraction.

Write the answer. $\frac{3}{\sqrt{1-9x^2}}$ .

State the restriction. The expression requires $|x| < \frac{1}{3}$ .

Exam Variation (Context / Parameter Shift): Let $f(x) = \arctan(x^2)$ . Find $f'(x)$ and $f'(1)$ .

Solution: With $u = x^2$ , the arctan formula gives $\frac{1}{1+x^4}$ , and the chain factor is $2x$ . So $f'(x) = \frac{2x}{1+x^4}$ and $f'(1) = \frac{2}{2} = 1$ . Note that the argument is squared inside the denominator and the chain factor stays outside the fraction; swapping those two placements is the standard error.

Common Traps & Failure Signals:

The unsubstituted argument (Signal: an $x^2$ where $9x^2$ belongs): the formula was written for the bare $x$ and never updated.
The sign of the partner (Signal: a positive arccos derivative): the pair was selected by shape rather than by name.
The swallowed chain factor (Signal: a factor of $1$ where the argument had a slope): the argument was substituted and its derivative was never multiplied in.

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

Pair Confusion $\rightarrow$ Say the function name before writing the formula.
Argument Shortcut: You copied the formula in its bare- $x$ form because that is the version memorised, and the substitution step felt redundant once the shape looked familiar.
Sign Drift: You kept the arctan sign while differentiating arccos.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The three formulas are judged in multiple choice; the two $\arctan$ appearances in the released sets are a limit value and a model whose derivative is supplied
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part; no scored part differentiates an inverse trigonometric function
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

A Linear Argument

What is the derivative of $f(x) = \arcsin(3x)$ ?

MCQ 2

A Composite Argument

Let $g(x) = \arctan(x^2)$ . What is $g'(x)$ ?

MCQ 3

The Sign of the Pair

What is the derivative of $h(x) = \arccos x$ ?

FRQ 4

An Inverse Tangent and a Tangent Line

Let $f(x) = \arctan(2x)$ .

(a) Find $f'(x)$ .

(b) Find the equation of the line tangent to the graph of $f$ at $x = \frac{1}{2}$ .

(c) Let $h(x) = f(x^2)$ . Find $h'(x)$ and evaluate $h'(1)$ .

(d) Explain why $f'(x)$ is positive for every real $x$ , and what that says about the graph of $f$ .

Cheatsheet

Inverse Trigonometric Derivatives

	Derivative
$\arcsin u$	$\dfrac{u'}{\sqrt{1-u^2}}$
$\arccos u$	$-\dfrac{u'}{\sqrt{1-u^2}}$
$\arctan u$	$\dfrac{u'}{1+u^2}$
$\operatorname{arccot} u$	$-\dfrac{u'}{1+u^2}$
$\operatorname{arcsec} u$	$\dfrac{u'}{\lvert u \rvert\sqrt{u^2-1}}$
$\operatorname{arccsc} u$	$-\dfrac{u'}{\lvert u \rvert\sqrt{u^2-1}}$

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Start Topic Practice Next Topic: 3.7 Derivatives of General Exponential and Logarithmic Functions

Unit 3: Differentiation: Composite, Implicit, and Inverse FunctionsDerivatives of Inverse Trigonometric Functions

Concept 4.136Built for May 2027 AP Exam

Derivatives of Inverse Trigonometric Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: $\arcsin$ , $\arccos$ , $\arctan$ , or $\operatorname{arcsec}$ applied to a bare $x$ or to a linear expression, usually inside a sum or a product.
What you’ll do: Select the correct formula from the pair, then multiply by the derivative of the argument.
Memory hook: Arcsin up, arccos down, arctan flat. The three shapes of the denominators follow that order.
Use it on the exam: For MCQs: The distractor with the sign of the partner function appears in every version, so confirm which of the pair you are differentiating before choosing. For FRQs: No released part scores these directly, but the reason to know them survives: when an inverse trig term sits inside a larger derivative it has to disappear when you differentiate it, and an answer still carrying one is unfinished.

Concept Review

Six Formulas, Three Pairs

SIX INVERSE TRIG SLOPES, THREE PAIRS

The Table Worth Memorising

Function	Derivative	Domain of the derivative
$\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\arccos x$	$-\dfrac{1}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\arctan x$	$\dfrac{1}{1+x^2}$	All real $x$
$\operatorname{arccot} x$	$-\dfrac{1}{1+x^2}$	All real $x$
$\operatorname{arcsec} x$	$\dfrac{1}{\lvert x \rvert\sqrt{x^2-1}}$	$\lvert x \rvert > 1$
$\operatorname{arccsc} x$	$-\dfrac{1}{\lvert x \rvert\sqrt{x^2-1}}$	$\lvert x \rvert > 1$

The absolute value in the secant and cosecant rows is not decoration. Without it the formula gives the wrong sign on the negative branch, and the exam tests that branch deliberately.

Composing With a Chain Factor

Common Misconception Check

Myth 1: " $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}[\arccos x]$ is the same." Truth: The arccos derivative carries a minus sign, and that sign propagates through every later term.
Myth 2: "The chain factor goes inside the square root." Truth: The chain factor multiplies the whole fraction, and the argument is squared inside the root.
Myth 3: " $\operatorname{arcsec}$ follows the same pattern as $\arcsin$ ." Truth: Its denominator carries an absolute value and a $\sqrt{x^2-1}$ , and the two branches have to be handled separately.

Topic-Specific Study Method

Pick the Pair, Then Chain

Exam disguise	What you see on the page	What your brain must translate	Required execution
A bare argument	$\arcsin x$	One formula, no chain factor	Write the formula and stop
A linear argument	$\arctan 4x$	The formula plus a constant factor	Substitute $4x$ , multiply by $4$
A composite argument	$\arcsin(x^2)$	Substitute the whole argument first	Square $x^2$ , then chain $2x$
A sum with a trig term	$\arcsin x + \sqrt{1-x^2}$	Two independent terms	Differentiate each, keep the signs
A secant form	$\operatorname{arcsec} x$	Absolute value and two branches	Handle each branch, or state the restriction

① Write the six formulas from memory three times in a row, each time grouping them into the three pairs. The pairing is what makes the signs survive an exam.

② Differentiate $\arcsin(3x)$ , $\arctan(3x)$ , and $\operatorname{arcsec}(3x)$ side by side and compare where the chain factor lands in each.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find $\frac{d}{dx}\left[\arcsin(3x)\right]$ .

Standard Execution:

Select the formula. The function is arcsin, so the denominator is $\sqrt{1-u^2}$ .

Substitute the argument. With $u = 3x$ the denominator becomes $\sqrt{1-9x^2}$ .

Attach the chain factor. The derivative of $3x$ is $3$ , which multiplies the fraction.

Write the answer. $\frac{3}{\sqrt{1-9x^2}}$ .

State the restriction. The expression requires $|x| < \frac{1}{3}$ .

Exam Variation (Context / Parameter Shift): Let $f(x) = \arctan(x^2)$ . Find $f'(x)$ and $f'(1)$ .

Solution: With $u = x^2$ , the arctan formula gives $\frac{1}{1+x^4}$ , and the chain factor is $2x$ . So $f'(x) = \frac{2x}{1+x^4}$ and $f'(1) = \frac{2}{2} = 1$ . Note that the argument is squared inside the denominator and the chain factor stays outside the fraction; swapping those two placements is the standard error.

Common Traps & Failure Signals:

The unsubstituted argument (Signal: an $x^2$ where $9x^2$ belongs): the formula was written for the bare $x$ and never updated.
The sign of the partner (Signal: a positive arccos derivative): the pair was selected by shape rather than by name.
The swallowed chain factor (Signal: a factor of $1$ where the argument had a slope): the argument was substituted and its derivative was never multiplied in.

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

Pair Confusion $\rightarrow$ Say the function name before writing the formula.
Argument Shortcut: You copied the formula in its bare- $x$ form because that is the version memorised, and the substitution step felt redundant once the shape looked familiar.
Sign Drift: You kept the arctan sign while differentiating arccos.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; The three formulas are judged in multiple choice; the two $\arctan$ appearances in the released sets are a limit value and a model whose derivative is supplied
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part; no scored part differentiates an inverse trigonometric function
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

A Linear Argument

What is the derivative of $f(x) = \arcsin(3x)$ ?

MCQ 2

A Composite Argument

Let $g(x) = \arctan(x^2)$ . What is $g'(x)$ ?

MCQ 3

The Sign of the Pair

What is the derivative of $h(x) = \arccos x$ ?

FRQ 4

An Inverse Tangent and a Tangent Line

Let $f(x) = \arctan(2x)$ .

(a) Find $f'(x)$ .

(b) Find the equation of the line tangent to the graph of $f$ at $x = \frac{1}{2}$ .

(c) Let $h(x) = f(x^2)$ . Find $h'(x)$ and evaluate $h'(1)$ .

(d) Explain why $f'(x)$ is positive for every real $x$ , and what that says about the graph of $f$ .

Cheatsheet

Inverse Trigonometric Derivatives

	Derivative
$\arcsin u$	$\dfrac{u'}{\sqrt{1-u^2}}$
$\arccos u$	$-\dfrac{u'}{\sqrt{1-u^2}}$
$\arctan u$	$\dfrac{u'}{1+u^2}$
$\operatorname{arccot} u$	$-\dfrac{u'}{1+u^2}$
$\operatorname{arcsec} u$	$\dfrac{u'}{\lvert u \rvert\sqrt{u^2-1}}$
$\operatorname{arccsc} u$	$-\dfrac{u'}{\lvert u \rvert\sqrt{u^2-1}}$

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