# Differentiating inverse trigonometric functions

> AP Calculus AB · Differentiation: Composite, Implicit, and Inverse Functions
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u3-differentiating-inverse-trigonometric-functions/

This sub-topic covers deriving derivative rules for inverse sine, cosine, and tangent, applying the chain rule to composite inverse trigonometric functions, solving tangent line problems, and identifying domains of differentiability for AP Calculus AB.

**Prerequisites:** [Implicit differentiation for inverse functions](https://www.owlsprep.com/study/ap-calculus-ab-u3-implicit-differentiation/); Derivatives of basic trigonometric functions; Chain rule for composite functions

## Learning objectives

- Derive derivative rules for arcsine, arccosine, and arctangent using implicit differentiation
- Differentiate composite inverse trigonometric functions using the chain rule
- Find tangent lines to inverse trigonometric functions at given points
- Identify domains of differentiability for inverse trigonometric functions

## Derivation of Basic Inverse Trigonometric Derivative Rules

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 $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, 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 $y \in [0, \pi]$ 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 $\frac{dy}{dx}$:
5. $$\frac{dy}{dx} = -\frac{1}{\sin y}$$
6. Use the Pythagorean identity to rewrite in terms of $x$. For $y \in [0, \pi]$, $\sin y \geq 0$, 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. $$\frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1$$

> **tip**
>
> On the AP exam, you do not need to re-derive these basic rules on a free-response question unless explicitly asked. You may cite them directly to save time.

## Differentiating Composite Inverse Trigonometric Functions

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: $\frac{dy}{dx} = f'(u(x)) \cdot u'(x)$, where $f$ is the outer inverse trig function.

The general chain rule forms for each core function are:

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

> **tip**
>
> The most common mistake here is forgetting to include $u'(x)$ in the numerator. Explicitly write down $u(x)$ and $u'(x)$ before applying the rule to avoid this error.

**Worked example:** Find the derivative of $f(x) = \arctan(3x^2 - 2x)$.

1. Identify outer and inner functions: outer function is $\arctan(u)$, inner function is $u(x) = 3x^2 - 2x$.
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. $$f'(x) = \frac{6x - 2}{1 + (3x^2 - 2x)^2} = \frac{6x - 2}{9x^4 - 12x^3 + 4x^2 + 1}$$

> **tip**
>
> Always check the domain of the derivative after finishing. For arcsine and arccosine composites, the derivative only exists where $|u(x)| < 1$, a common point tested in multiple-choice domain questions.

## Tangent Line Problems for Inverse Trigonometric Functions

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 $y = \arcsin\left(\frac{x}{2}\right)$ at $x = 1$.

1. Find the y-coordinate of the tangency point. $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$ (we do not use $\frac{5\pi}{6}$ because arcsine is restricted to $[-\frac{\pi}{2}, \frac{\pi}{2}]$), so the point is $\left(1, \frac{\pi}{6}\right)$.
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. $$y = \frac{\sqrt{3}}{3}x - \frac{\sqrt{3}}{3} + \frac{\pi}{6}$$

> **tip**
>
> Always confirm your y-value falls within the inverse trig function's restricted range before using it in the tangent line equation.

## Mixed AP-Style Worked Examples

**Worked example:** What is the derivative of $f(x) = x \arccos(2x)$?

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 $\arccos(2x)$, since $\frac{d}{dx}[x] = 1$. For the second term, $u(x) = 2x$, 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. $f'(x) = \arccos(2x) - \frac{2x}{\sqrt{1 - 4x^2}}$

**Worked example:** Let $f(x) = \arctan(4x)$. (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, $u(x) = 4x$, $u'(x) = 4$:
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. $$\frac{4}{1 + 16x^2} = \frac{1}{2} \implies 8 = 1 + 16x^2 \implies 16x^2 = 7 \implies x = \pm \frac{\sqrt{7}}{4}$$

## Common pitfalls

- **Wrong:** Writing $\frac{d}{dx} \arcsin(2x) = \frac{1}{\sqrt{1 - 4x^2}}$ (omitting the factor of 2 from the chain rule)
  - Why it fails: Students memorize the basic rule for $\arcsin x$ and forget composite functions require multiplying by the derivative of the inner function.
  - Correct: 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:** Writing $\frac{d}{dx} \arcsin x = \pm \frac{1}{\sqrt{1 - x^2}}$ (keeping the $\pm$ from the square root)
  - Why it fails: Students forget the range restriction on inverse sine that guarantees $\cos y$ is non-negative, so they leave an ambiguous sign.
  - Correct: Always use the positive root for the derivative of $\arcsin x$ and the fixed negative root for $\arccos x$, never $\pm$.
- **Wrong:** Claiming the derivative of $f(x) = \arccos(x^3)$ exists at $x = 1$
  - Why it fails: 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: Always check that $|u(x)| < 1$ for arcsine and arccosine derivatives when asked about differentiability at a point.
- **Wrong:** Writing $\frac{d}{dx} \arctan(x^2) = \frac{2x}{1 - x^4}$
  - Why it fails: Students confuse the denominator of the arctangent derivative with the denominator of arcsine, swapping the plus sign for a minus.
  - Correct: Remember: sine/cosine inverses have $1 - u^2$ in the denominator, arctangent has $1 + u^2$.
- **Wrong:** Stopping at $\frac{d}{dx} \arcsin x = \frac{1}{\cos(\arcsin x)}$ instead of simplifying
  - Why it fails: Students remember the general inverse function derivative rule but forget to simplify to the standard form in x required on the exam.
  - Correct: Always simplify inverse trig derivatives using the Pythagorean identity to get an expression in x only, with no inverse trig remaining.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Basic arcsine derivative | $\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1 - x^2}}$ | Valid for $\|x\| < 1$; undefined at endpoints $x = \pm 1$ |
| Basic arccosine derivative | $\frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1 - x^2}}$ | Valid for $\|x\| < 1$; always negative, opposite sign of arcsine |
| Basic arctangent derivative | $\frac{d}{dx} \arctan x = \frac{1}{1 + x^2}$ | Valid for all real $x$; denominator has a plus sign, not minus |
| Chain rule: arcsine composite | $\frac{d}{dx} \arcsin(u(x)) = \frac{u'(x)}{\sqrt{1 - [u(x)]^2}}$ | Always multiply by $u'(x)$; valid where $\|u(x)\| < 1$ |
| Chain rule: arccosine composite | $\frac{d}{dx} \arccos(u(x)) = -\frac{u'(x)}{\sqrt{1 - [u(x)]^2}}$ | Do not forget the negative sign on the numerator |
| Chain rule: arctangent composite | $\frac{d}{dx} \arctan(u(x)) = \frac{u'(x)}{1 + [u(x)]^2}$ | 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 |

## 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.

- [Implicit differentiation](https://www.owlsprep.com/study/ap-calculus-ab-u3-implicit-differentiation/)
- [Differentiating inverse functions](https://www.owlsprep.com/study/ap-calculus-ab-u3-differentiating-inverse-functions/)
- [Selecting procedures for calculating derivatives](https://www.owlsprep.com/study/ap-calculus-ab-u3-selecting-procedures-for-calculating-derivatives/)

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