# Differentiating inverse trigonometric functions

> AP Calculus BC · Unit 3: Differentiation: Composite, Implicit, and Inverse Functions
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u3-differentiating-inverse-trigonometric-functions/

This module covers deriving derivatives of all six inverse trigonometric functions, applying the chain rule to composite inverse trig functions, and combining inverse trig differentiation with product/quotient rule for AP Calculus BC exam prep.

**Prerequisites:** [Implicit differentiation for inverse functions](https://www.owlsprep.com/study/ap-calculus-bc-u3-implicit-differentiation/); [Chain rule for composite functions](https://www.owlsprep.com/study/ap-calculus-bc-u3-chain-rule-composite-functions/); Domain and range of principal branch inverse trigonometric functions

## Learning objectives

- Derive derivative formulas for inverse trigonometric functions using implicit differentiation
- Differentiate composite inverse trigonometric functions using the chain rule
- Differentiate combinations of inverse trig functions with product/quotient rule
- Identify and avoid common mistakes when working with inverse trig derivatives

## Deriving Derivatives of Basic Inverse Trigonometric Functions

All derivative formulas for inverse trigonometric functions can be derived from the definition of inverse functions and implicit differentiation. The process follows the same steps for all six functions: rewrite the inverse trig equation as a regular trig equation, differentiate implicitly, solve for $\frac{dy}{dx}$, use a Pythagorean identity to rewrite in terms of $x$, and adjust the sign to match the principal range of the inverse function.

**Worked example:** 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 $x$:
3. $$\frac{d}{dx}(\tan y) = \frac{d}{dx}(x) \implies \sec^2 y \cdot \frac{dy}{dx} = 1$$
4. Solve for $\frac{dy}{dx}$:
5. $$\frac{dy}{dx} = \frac{1}{\sec^2 y}$$
6. Use the Pythagorean identity $1 + \tan^2 y = \sec^2 y$, and substitute $\tan y = x$:
7. $$\frac{dy}{dx} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$$
8. 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:
9. $$\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2}$$

> **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.

## Differentiating Composite Inverse Trigonometric Functions

Virtually all AP exam problems on this topic involve composite inverse trigonometric functions, where the argument of the inverse trig function is another function of $x$. To differentiate these, 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 $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

> **note**
>
> Always substitute the entire inner function $u(x)$ into the inverse trig derivative formula before multiplying by $u'(x)$.

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

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

> **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.

## Differentiating Combinations of Inverse Trigonometric Functions

AP problems often require differentiating combinations of inverse trig functions with other function types (polynomials, exponentials, logarithms) using the product rule or quotient rule. The process is identical to differentiating any other combination: apply the product or quotient rule first, then compute the derivative of each inverse trig term using the inverse trig rule and chain rule.

**Worked example:** 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:
2. $$f'(x) = \frac{d}{dx}(e^x) \cdot \arcsin(x) + e^x \cdot \frac{d}{dx}(\arcsin(x))$$
3. Substitute the derivatives of each term:
4. $$f'(x) = e^x \arcsin(x) + \frac{e^x}{\sqrt{1 - x^2}}$$
5. Evaluate $f'(x)$ at $x = 0$: $\arcsin(0) = 0$, $e^0 = 1$, $\sqrt{1 - 0^2} = 1$.
6. Simplify to get the slope:
7. $$f'(0) = (1 \cdot 0) + \frac{1}{1} = 1$$

> **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.

## AP-Style Practice

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. 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) $\frac{2}{e^{2x}\sqrt{e^{4x} - 1}}$

   *Why:* We use the chain rule: $u = e^{2x}$, $u' = 2e^{2x}$, $|u| = e^{2x}$ (always positive). The formula gives $\frac{2e^{2x}}{e^{2x}\sqrt{e^{4x} - 1}} = \frac{2}{\sqrt{e^{4x} - 1}}$ after cancellation.

**Worked example:** A camera is positioned 100 meters horizontally from a rocket launch pad. The rocket rises straight up at a constant speed of 50 m/s. Let $\theta$ be the angle between the camera's line of sight and the horizontal. Find the rate of change of $\theta$ when the rocket is 100 meters above the ground.

1. Let $h$ = height of the rocket, so $\frac{dh}{dt} = 50$ m/s, and horizontal distance is constant at 100 m. By trigonometry: $\tan\theta = \frac{h}{100}$, so $\theta = \arctan\left(\frac{h}{100}\right)$.
2. Differentiate both sides with respect to time $t$, using the chain rule:
3. $$\frac{d\theta}{dt} = \frac{1}{1 + \left(\frac{h}{100}\right)^2} \cdot \frac{1}{100} \cdot \frac{dh}{dt}$$
4. Substitute $h = 100$ m and $\frac{dh}{dt} = 50$ m/s:
5. $$\frac{d\theta}{dt} = \frac{1}{1 + (1)^2} \cdot \frac{1}{100} \cdot 50 = 0.25$$
6. The angle is increasing at a rate of 0.25 radians per second.

## Common pitfalls

- **Wrong:** Differentiating $y = \arcsin(5x)$ and writing $\frac{dy}{dx} = \frac{1}{\sqrt{1 - (5x)^2}}$, omitting the chain rule term.
  - Why it fails: Students remember the basic derivative of $\arcsin x$ but forget the inner function $5x$ has a non-zero derivative, incorrectly assuming $u' = 1$.
  - Correct: Always explicitly label the inner function $u(x)$ and write down $u'$ before applying the inverse trig derivative rule, so you do not forget to include it as a factor.
- **Wrong:** Writing the derivative of $\arccos(x)$ as $\frac{1}{\sqrt{1 - x^2}}$, missing the negative sign.
  - Why it fails: Students mix up the derivatives of $\arcsin x$ and $\arccos x$, which only differ by a sign.
  - Correct: 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:** Differentiating $y = \text{arcsec}(3x)$ and writing $\frac{dy}{dx} = \frac{1}{3x\sqrt{9x^2 - 1}}$, omitting the absolute value around $3x$.
  - Why it fails: Students forget the derivative of $\text{arcsec}$ requires an absolute value from domain restrictions.
  - Correct: Always add an absolute value around the $u$ term when writing the derivative of $\text{arcsec}$ or $\text{arccsc}$, per the standard formula.
- **Wrong:** 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 it fails: Students rush and only substitute the inner function into the chain rule term, not into the inverse trig derivative formula.
  - Correct: After identifying $u(x)$, substitute $u$ into the inverse trig derivative formula before multiplying by $u'$.
- **Wrong:** Claiming $\frac{d}{dx}[\arcsin(\sin 2x)] = 2$ for all real $x$.
  - Why it fails: 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: Check the domain where the identity holds before differentiating, and only simplify if the problem specifies the interval of interest.

## Cheatsheet

| Inverse Trig Function | Derivative Formula | Notes |
| --- | --- | --- |
| $\arcsin(u)$ | $\displaystyle \frac{u'}{\sqrt{1 - u^2}}$ | Valid for $-1 < u < 1$, always positive |
| $\arccos(u)$ | $\displaystyle -\frac{u'}{\sqrt{1 - u^2}}$ | Valid for $-1 < u < 1$, always negative |
| $\arctan(u)$ | $\displaystyle \frac{u'}{1 + u^2}$ | Valid for all real $u$, always positive |
| $\text{arccot}(u)$ | $\displaystyle -\frac{u'}{1 + u^2}$ | Valid for all real $u$, always negative |
| $\text{arcsec}(u)$ | $\displaystyle \frac{u'}{\|u\|\sqrt{u^2 - 1}}$ | Valid for $\|u\| > 1$, always positive |
| $\text{arccsc}(u)$ | $\displaystyle -\frac{u'}{\|u\|\sqrt{u^2 - 1}}$ | Valid for $\|u\| > 1$, always negative |
| Chain Rule | If $y = f(g(x))$, $y' = f'(g(x)) \cdot g'(x)$ | Never forget the $g'(x)$ term |
| Product/Quotient | Apply rule first, then substitute derivatives | Set up the rule before computing inverse trig derivatives |

## 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 that an integrand matches the derivative of an inverse trig function is the core skill for these integration problems, which are commonly tested on both multiple-choice and free-response 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](https://www.owlsprep.com/study/ap-calculus-bc-u3-implicit-differentiation/)
- [Selecting procedures for calculating derivatives](https://www.owlsprep.com/study/ap-calculus-bc-u3-selecting-procedures-for-calculating-derivatives/)
- [Calculating higher-order derivatives](https://www.owlsprep.com/study/ap-calculus-bc-u3-calculating-higher-order-derivatives/)

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