Differentiating inverse trigonometric functions

AP Calculus BC· AP Calculus BC CED — Differentiation: Composite, Implicit, and Inverse Functions· 14 min read

1. Deriving Derivatives of Basic Inverse Trigonometric Functions★★☆☆☆⏱ 4 min

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 $\frac{d y}{d x}$ , 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 $- \frac{π}{2} < y < \frac{π}{2}$ .
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 $\frac{d y}{d x}$ :
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
$\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2}$

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

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 $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ .

📐 Worked Example

Find $\frac{d y}{d x}$ for $y = arccos (2 x - 1)$ .

1
Identify the outer and inner functions: $y = arccos (u)$ , $u (x) = 2 x - 1$ , so $u^{'} (x) = 2$ .
2
Recall the derivative rule for $arccos (u)$ : $\frac{d}{d x} [arccos (u)] = - \frac{u ^{'}}{1 - u ^{2}}$ .
3
Substitute $u = 2 x - 1$ 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
$\frac{dy}{dx} = -\frac{2}{\sqrt{4x(1 - x)}} = -\frac{2}{2\sqrt{x(1 - x)}} = -\frac{1}{\sqrt{x(1 - x)}}$

3. Differentiating Combinations of Inverse Trigonometric Functions★★★☆☆⏱ 3 min

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 $x = 0$ : $arcsin (0) = 0$ , $e^{0} = 1$ , $1 - 0^{2} = 1$ .
6
Simplify to get the slope:
7
$f'(0) = (1 \cdot 0) + \frac{1}{1} = 1$

4. AP-Style Practice★★★★☆⏱ 3 min

✓ Quick check

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

Which of the following is equal to $\frac{d}{d x} [arcsec (e^{2 x})]$ ?
- A) $\frac{1}{e ^{2 x} e ^{4 x} - 1}$
- B) $\frac{2}{e ^{4 x} - 1}$
- C) $\frac{2 e ^{2 x}}{e ^{4 x} - 1}$
- D) $\frac{2}{e ^{2 x} e ^{4 x} - 1}$
Reveal answer
B —
We use the chain rule: $u = e^{2 x}$ , $u^{'} = 2 e^{2 x}$ , $∣ u ∣ = e^{2 x}$ (always positive). The formula gives $\frac{2 e ^{2 x}}{e ^{2 x} e ^{4 x} - 1} = \frac{2}{e ^{4 x} - 1}$ 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 $θ$ be the angle between the camera's line of sight and the horizontal. Find the rate of change of $θ$ when the rocket is 100 meters above the ground.

1
Let $h$ = height of the rocket, so $\frac{d h}{d t} = 50$ m/s, and horizontal distance is constant at 100 m. By trigonometry: $tan θ = \frac{h}{100}$ , so $θ = arctan (\frac{h}{100})$ .
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 $\frac{d h}{d t} = 50$ 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.

5. Common Pitfalls

Wrong move:

Differentiating $y = arcsin (5 x)$ and writing $\frac{d y}{d x} = \frac{1}{1 - ( 5 x ) ^{2}}$ , omitting the chain rule term.

Why:

Students remember the basic derivative of $arcsin x$ but forget the inner function $5 x$ has a non-zero derivative, incorrectly assuming $u^{'} = 1$ .

Correct move:

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 move:

Writing the derivative of $arccos (x)$ as $\frac{1}{1 - x ^{2}}$ , missing the negative sign.

Why:

Students mix up the derivatives of $arcsin x$ and $arccos x$ , which only differ by a sign.

Correct move:

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 move:

Differentiating $y = arcsec (3 x)$ and writing $\frac{d y}{d x} = \frac{1}{3 x 9 x ^{2} - 1}$ , omitting the absolute value around $3 x$ .

Why:

Students forget the derivative of $arcsec$ requires an absolute value from domain restrictions.

Correct move:

Always add an absolute value around the $u$ term when writing the derivative of $arcsec$ or $arccsc$ , per the standard formula.

Wrong move:

Differentiating $y = arctan (x^{2})$ and writing $\frac{d y}{d x} = \frac{2 x}{1 + x ^{2}}$ , substituting $x$ instead of $x^{2}$ into the denominator.

Why:

Students rush and only substitute the inner function into the chain rule term, not into the inverse trig derivative formula.

Correct move:

After identifying $u (x)$ , substitute $u$ into the inverse trig derivative formula before multiplying by $u^{'}$ .

Wrong move:

Claiming $\frac{d}{d x} [arcsin (sin 2 x)] = 2$ for all real $x$ .

Why:

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 move:

Check the domain where the identity holds before differentiating, and only simplify if the problem specifies the interval of interest.

6. Quick Reference Cheatsheet

Inverse Trig Function	Derivative Formula	Notes
$arcsin (u)$	$\frac{u ^{'}}{1 - u ^{2}}$	Valid for $- 1 < u < 1$ , always positive
$arccos (u)$	$- \frac{u ^{'}}{1 - u ^{2}}$	Valid for $- 1 < u < 1$ , always negative
$arctan (u)$	$\frac{u ^{'}}{1 + u ^{2}}$	Valid for all real $u$ , always positive
$arccot (u)$	$- \frac{u ^{'}}{1 + u ^{2}}$	Valid for all real $u$ , always negative
$arcsec (u)$	$\frac{u ^{'}}{∣ u ∣ u ^{2} - 1}$	Valid for $∣ u ∣ > 1$ , always positive
$arccsc (u)$	$- \frac{u ^{'}}{∣ u ∣ u ^{2} - 1}$	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

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
Differentiate composite inverse trig function
2022 · FRQ
Related rates with inverse trig

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 $\frac{1}{a ^{2} + x ^{2}}$ , $\frac{1}{a ^{2} - x ^{2}}$ , and $\frac{1}{x x ^{2} - a ^{2}}$ . 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.