Differentiating inverse trigonometric functions

AP Calculus AB· AP Calculus AB CED — Unit 3 Differentiation· 14 min read

1. Derivation of Basic Inverse Trigonometric Derivative Rules★★☆☆☆⏱ 4 min

All derivative rules for inverse trigonometric functions can be derived using implicit differentiation, leveraging the inverse function definition: if $y = f^{- 1} (x)$ , then $f (y) = x$ for the restricted range of the inverse function.

For $y = arcsin x$ , we start with $sin y = x$ , where $y \in [- \frac{π}{2}, \frac{π}{2}]$ , so $cos y \geq 0$ . Differentiating implicitly with respect to $x$ gives:

\cos y \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\cos y}

Using the Pythagorean identity $cos^{2} y = 1 - sin^{2} y = 1 - x^{2}$ , we take the positive root to get the final rule:

\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1

Following the same process gives the rules for arccosine and arctangent:

\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1 \\ \frac{d}{dx}\arctan x = \frac{1}{1 + x^2}, \quad \text{all real } x

📐 Worked Example

Derive the derivative rule for $f (x) = arccos x$ using implicit differentiation.

1
Start with the inverse function definition: let $y = arccos x$ , so $cos y = x$ , where $y \in [0, π]$ by definition of the restricted inverse cosine function.
2
Differentiate both sides with respect to $x$ , applying the chain rule to the left-hand side:
3
$-\sin y \cdot \frac{dy}{dx} = 1$
4
Isolate $\frac{d y}{d x}$ :
5
$\frac{dy}{dx} = -\frac{1}{\sin y}$
6
Use the Pythagorean identity to rewrite in terms of $x$ . For $y \in [0, π]$ , $sin y \geq 0$ , so:
7
$\sin^2 y = 1 - \cos^2 y = 1 - x^2 \implies \sin y = \sqrt{1 - x^2}$
8
Substitute back to get the final rule:
9
$\frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1$

2. Differentiating Composite Inverse Trigonometric Functions★★★☆☆⏱ 4 min

Nearly all AP exam problems involve composite inverse trigonometric functions of the form $y = arcsin (u (x))$ , $y = arccos (u (x))$ , or $y = arctan (u (x))$ , where $u (x)$ is a non-constant differentiable function. To differentiate these, apply the chain rule: $\frac{d y}{d x} = f^{'} (u (x)) \cdot u^{'} (x)$ , where $f$ is the outer inverse trig function.

The general chain rule forms for each core function are:

\frac{d}{dx}\arcsin(u(x)) = \frac{u'(x)}{\sqrt{1 - [u(x)]^2}}, \quad |u(x)| < 1 \\ \frac{d}{dx}\arccos(u(x)) = \frac{-u'(x)}{\sqrt{1 - [u(x)]^2}}, \quad |u(x)| < 1 \\ \frac{d}{dx}\arctan(u(x)) = \frac{u'(x)}{1 + [u(x)]^2}, \quad \text{all real } x

📐 Worked Example

Find the derivative of $f (x) = arctan (3 x^{2} - 2 x)$ .

1
Identify outer and inner functions: outer function is $arctan (u)$ , inner function is $u (x) = 3 x^{2} - 2 x$ .
2
Calculate the derivative of the inner function:
3
$u'(x) = 6x - 2$
4
Apply the chain rule for arctangent:
5
$f'(x) = \frac{u'(x)}{1 + [u(x)]^2}$
6
Substitute and simplify to get the final result:
7
$f'(x) = \frac{6x - 2}{1 + (3x^2 - 2x)^2} = \frac{6x - 2}{9x^4 - 12x^3 + 4x^2 + 1}$

3. Tangent Line Problems for Inverse Trigonometric Functions★★★☆☆⏱ 3 min

A common AP exam application of inverse trig derivatives is finding the equation of a tangent line to an inverse trigonometric function at a given point. This requires three key steps: (1) calculating the correct y-coordinate using the inverse's restricted range, (2) finding the derivative at the point for slope, (3) writing the tangent line in point-slope form, which is preferred unless another form is requested.

📐 Worked Example

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

1
Find the y-coordinate of the tangency point. $arcsin (\frac{1}{2}) = \frac{π}{6}$ (we do not use $\frac{5 π}{6}$ because arcsine is restricted to $[- \frac{π}{2}, \frac{π}{2}]$ ), so the point is $(1, \frac{π}{6})$ .
2
Calculate the derivative using the chain rule: $u (x) = \frac{x}{2}$ , so $u^{'} (x) = \frac{1}{2}$ :
3
$y' = \frac{\frac{1}{2}}{\sqrt{1 - \left(\frac{x}{2}\right)^2}} = \frac{1}{\sqrt{4 - x^2}}$
4
Find the slope at $x = 1$ :
5
$y'(1) = \frac{1}{\sqrt{4 - 1}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
6
Write the tangent line in point-slope form:
7
$y - \frac{\pi}{6} = \frac{\sqrt{3}}{3}(x - 1)$
8
Converting to slope-intercept form (if requested) gives:
9
$y = \frac{\sqrt{3}}{3}x - \frac{\sqrt{3}}{3} + \frac{\pi}{6}$

4. Mixed AP-Style Worked Examples★★★★☆⏱ 3 min

📐 Worked Example

What is the derivative of $f (x) = x arccos (2 x)$ ?

1
This problem requires the product rule and chain rule for inverse trigonometric functions. By the product rule:
2
$f'(x) = \frac{d}{dx}[x] \cdot \arccos(2x) + x \cdot \frac{d}{dx}[\arccos(2x)]$
3
The first term simplifies to $arccos (2 x)$ , since $\frac{d}{d x} [x] = 1$ . For the second term, $u (x) = 2 x$ , so $u^{'} (x) = 2$ :
4
$\frac{d}{dx}\arccos(2x) = \frac{-2}{\sqrt{1 - 4x^2}}$
5
Multiply by $x$ and combine terms to get the final result:
6
$f^{'} (x) = arccos (2 x) - \frac{2 x}{1 - 4 x ^{2}}$

📐 Worked Example

Let $f (x) = arctan (4 x)$ . (a) Find $f^{'} (x)$ ; (b) Find the slope of the tangent line at $x = \frac{1}{4}$ ; (c) Find all $x$ where the slope equals $\frac{1}{2}$ .

1
(a) Use the chain rule for composite arctangent, $u (x) = 4 x$ , $u^{'} (x) = 4$ :
2
$f'(x) = \frac{4}{1 + (4x)^2} = \frac{4}{1 + 16x^2}$
3
(b) Substitute $x = \frac{1}{4}$ into $f^{'} (x)$ to get the slope:
4
$f'\left(\frac{1}{4}\right) = \frac{4}{1 + 16\left(\frac{1}{4}\right)^2} = \frac{4}{1 + 1} = 2$
5
(c) Set $f^{'} (x) = \frac{1}{2}$ and solve for $x$ :
6
$\frac{4}{1 + 16x^2} = \frac{1}{2} \implies 8 = 1 + 16x^2 \implies 16x^2 = 7 \implies x = \pm \frac{\sqrt{7}}{4}$

5. Common Pitfalls

Wrong move:

Writing $\frac{d}{d x} arcsin (2 x) = \frac{1}{1 - 4 x ^{2}}$ (omitting the factor of 2 from the chain rule)

Why:

Students memorize the basic rule for $arcsin x$ and forget composite functions require multiplying by the derivative of the inner function.

Correct move:

Always explicitly label the inner function $u (x)$ and its derivative $u^{'} (x)$ before applying the inverse trig derivative rule, so you do not forget $u^{'} (x)$ in the numerator.

Wrong move:

Writing $\frac{d}{d x} arcsin x = \pm \frac{1}{1 - x ^{2}}$ (keeping the $\pm$ from the square root)

Why:

Students forget the range restriction on inverse sine that guarantees $cos y$ is non-negative, so they leave an ambiguous sign.

Correct move:

Always use the positive root for the derivative of $arcsin x$ and the fixed negative root for $arccos x$ , never $\pm$ .

Wrong move:

Claiming the derivative of $f (x) = arccos (x^{3})$ exists at $x = 1$

Why:

Students forget the derivative only exists where $∣ u (x) ∣ < 1$ , not $∣ u (x) ∣ \leq 1$ , because the derivative is undefined at the endpoints of the domain.

Correct move:

Always check that $∣ u (x) ∣ < 1$ for arcsine and arccosine derivatives when asked about differentiability at a point.

Wrong move:

Writing $\frac{d}{d x} arctan (x^{2}) = \frac{2 x}{1 - x ^{4}}$

Why:

Students confuse the denominator of the arctangent derivative with the denominator of arcsine, swapping the plus sign for a minus.

Correct move:

Remember: sine/cosine inverses have $1 - u^{2}$ in the denominator, arctangent has $1 + u^{2}$ .

Wrong move:

Stopping at $\frac{d}{d x} arcsin x = \frac{1}{c o s ( a r c s i n x )}$ instead of simplifying

Why:

Students remember the general inverse function derivative rule but forget to simplify to the standard form in x required on the exam.

Correct move:

Always simplify inverse trig derivatives using the Pythagorean identity to get an expression in x only, with no inverse trig remaining.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Basic arcsine derivative	$\frac{d}{d x} arcsin x = \frac{1}{1 - x ^{2}}$	Valid for $∣ x ∣ < 1$ ; undefined at endpoints $x = \pm 1$
Basic arccosine derivative	$\frac{d}{d x} arccos x = - \frac{1}{1 - x ^{2}}$	Valid for $∣ x ∣ < 1$ ; always negative, opposite sign of arcsine
Basic arctangent derivative	$\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}$	Valid for all real $x$ ; denominator has a plus sign, not minus
Chain rule: arcsine composite	$\frac{d}{d x} arcsin (u (x)) = \frac{u ^{'} ( x )}{1 - [ u ( x ) ] ^{2}}$	Always multiply by $u^{'} (x)$ ; valid where $∣ u (x) ∣ < 1$
Chain rule: arccosine composite	$\frac{d}{d x} arccos (u (x)) = - \frac{u ^{'} ( x )}{1 - [ u ( x ) ] ^{2}}$	Do not forget the negative sign on the numerator
Chain rule: arctangent composite	$\frac{d}{d x} arctan (u (x)) = \frac{u ^{'} ( x )}{1 + [ u ( x ) ] ^{2}}$	Valid for all x where $u (x)$ is differentiable
Tangent line at $x = a$	$y - f (a) = f^{'} (a) (x - a)$	Always confirm $f (a)$ uses the correct restricted range of the inverse function

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
Chain rule for composite arcsine
2022 · FRQ
Tangent line to arctangent function
2021 · MCQ
Domain of inverse trig derivative

What's Next

Mastering derivatives of inverse trigonometric functions is a critical prerequisite for recognizing inverse trigonometric antiderivatives, which you will learn in Unit 6 of the AP Calculus AB syllabus. Without these derivative rules memorized, you will struggle to match integrands to their antiderivatives on integration problems, which make up a much larger share of the AP exam score. This topic also forms the foundation for inverse trigonometric terms in related rates and optimization problems later in Unit 4, and is commonly combined with implicit differentiation and product/quotient rule questions in multiple-choice sections. Building fluency with these rules now will save you time on more complex topics later in the course.