Unit 3 · Topic 3.4 · about 30 minutes · 7 parts · 11 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 3.4: Differentiating Inverse Trigonometric Functions

Q: Is -1 x the same as x?

Yes. Both name inverse sine. The -1 is not an exponent, so neither one means 1 x.

Q: Should I memorize these derivatives or work them out each time?

Know the three results well enough to use them at once. The derivation is your backup when you doubt a sign or a root: write y = x (or y = x, or y = x), differentiate both sides, and replace the trigonometric function of y with an expression in x.

Q: Why does the derivative of x not exist at x = 1?

The formula 11 - x2 divides by 0 there. On the graph, inverse sine gets steeper and steeper as x approaches 1, and sine itself has slope 0 at 2, whose reciprocal does not exist.

Q: Why are the derivatives of x and x opposites?

For -1 x 1, x + x = 2, a constant. A constant has derivative 0, so the two derivatives add to 0.

Q: What about inverse cotangent, inverse secant and inverse cosecant?

The same two methods find their derivatives. Each one depends on the interval its trigonometric function is restricted to, so write the definition with its interval before you differentiate.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 3: 5 to 10% of the multiple-choice section

What you'll be able to do

Calculate the derivatives of inverse sine, inverse cosine and inverse tangent, alone or with an inside function, from the chain rule and the definition of an inverse function.

Key vocabulary

Term	What it means	The exam trap
Inverse function	A function $g$ that undoes a one-to-one function $f$ , so that $g (f (x)) = x$ and $f (g (x)) = x$ on their domains.	A trigonometric function has an inverse only after its domain is restricted to an interval where it takes each value once.
Inverse sine, $\arcsin x$	For $- 1 \leq x \leq 1$ , the angle $y$ with $- \frac{π}{2} \leq y \leq \frac{π}{2}$ whose sine is $x$ ; also written $\sin^{- 1} x$ .	In $\sin^{- 1} x$ the $- 1$ marks an inverse, not a power, so $\sin^{- 1} x$ is not $\frac{1}{\sin x}$ , which is $\csc x$ .
Inverse cosine, $\arccos x$	For $- 1 \leq x \leq 1$ , the angle $y$ with $0 \leq y \leq π$ whose cosine is $x$ ; also written $\cos^{- 1} x$ .	In $\cos^{- 1} x$ the $- 1$ marks an inverse, so $\cos^{- 1} x$ is not $\sec x$ .
Inverse tangent, $\arctan x$	For every real $x$ , the angle $y$ with $- \frac{π}{2} < y < \frac{π}{2}$ whose tangent is $x$ ; also written $\tan^{- 1} x$ .	The endpoints $- \frac{π}{2}$ and $\frac{π}{2}$ are left out, because tangent has no value there.

Graph of y = sin x from -pi/2 to pi/2 and its reflection across the dashed line y = x, the graph of y = arcsin x from -1 to 1. A tangent line touches sine at (0.927, 0.8), and another touches arcsin at the reflected point (0.8, 0.927).

Part 1 Think first Raise the ramp a little

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A ramp runs 1 meter along the ground and rises $x$ meters, so the angle $θ$ it makes with the ground, in radians, has $\tan θ = x$ . With a rise of 1 meter the angle is $\frac{π}{4}$ . A builder raises the top of the ramp by 0.01 meter. About how much does the angle grow?

In short. The derivatives of the inverse trigonometric functions come from the definition of an inverse function with the chain rule, or from the formula for the derivative of an inverse: $\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}}$ , $\frac{d}{d x} \arccos x = - \frac{1}{\sqrt{1 - x^{2}}}$ and $\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}$ .

Each inverse undoes its trigonometric function on one restricted interval, and that interval decides the sign of any square root in the derivative.
With an inside function $u$ , multiply by $\frac{d u}{d x}$ , and square all of $u$ in the denominator.
The derivatives of $\arcsin x$ and $\arccos x$ exist only for $- 1 < x < 1$ ; the derivative of $\arctan x$ exists for every real $x$ .

Part 2 Learn it Three short ideas, each checked as you go

4a. Inverse sine, and its derivative from the chain rule

Sine takes every value from $- 1$ to 1 many times, so it has no inverse on its whole domain. Keep only the inputs from $- \frac{π}{2}$ to $\frac{π}{2}$ , where sine rises from $- 1$ to 1 and takes each value once. On that interval sine has an inverse, the inverse sine, written $\arcsin x$ or $\sin^{- 1} x$ :

y = \arcsin x means \sin y = x and - \frac{π}{2} \leq y \leq \frac{π}{2},

for $- 1 \leq x \leq 1$ .

To find its derivative, differentiate both sides of $\sin y = x$ with respect to $x$ . Because $y$ depends on $x$ , the chain rule puts a $\frac{d y}{d x}$ on the left:

\cos y \cdot \frac{d y}{d x} = 1, so \frac{d y}{d x} = \frac{1}{\cos y} .

Now write $\cos y$ in terms of $x$ . From $\sin^{2} y + \cos^{2} y = 1$ , $\cos^{2} y = 1 - x^{2}$ . The interval settles the sign of the root: $y$ is between $- \frac{π}{2}$ and $\frac{π}{2}$ , where cosine is never negative, so $\cos y = \sqrt{1 - x^{2}}$ and

\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}} for - 1 < x < 1.

At $x = 1$ and $x = - 1$ the denominator is 0, so the derivative does not exist there.

Quick check · Decide and justify

No calculator. A classmate differentiates both sides of $\sin y = x$ , where $y = \arcsin x$ , and gets $\frac{d y}{d x} = \frac{1}{\cos y}$ . Since $\cos^{2} y = 1 - x^{2}$ , the classmate says $\cos y$ could be $- \sqrt{1 - x^{2}}$ , so the derivative of $\arcsin x$ is negative for some $x$ with $- 1 < x < 1$ . Is the classmate right?

Which reason supports your answer?

4b. Inverse tangent and inverse cosine, from the inverse function formula

If $g$ is the inverse of $f$ , then $g^{'} (x) = \frac{1}{f^{'} (g (x))}$ wherever $f^{'} (g (x)) \neq 0$ . The inverse's slope is the reciprocal of the original's slope at the reflected point. In the graph at the top, the tangent to sine at $(0.927, 0.8)$ has slope 0.6, and the tangent to inverse sine at the reflected point $(0.8, 0.927)$ has slope $\frac{1}{0.6}$ , about 1.67.

Tangent, kept to $- \frac{π}{2} < y < \frac{π}{2}$ , has the inverse $y = \arctan x$ , defined for every real $x$ . The derivative of $\tan y$ is $\sec^{2} y$ , so

\frac{d}{d x} \arctan x = \frac{1}{\sec^{2} (\arctan x)} = \frac{1}{1 + x^{2}},

because $\sec^{2} y = 1 + \tan^{2} y$ and $\tan (\arctan x) = x$ .

Cosine, kept to $0 \leq y \leq π$ , has the inverse $y = \arccos x$ for $- 1 \leq x \leq 1$ . The derivative of $\cos y$ is $- \sin y$ , so the formula gives $\frac{1}{- \sin y}$ . On $[0, π]$ sine is never negative, so $\sin y = \sqrt{1 - x^{2}}$ and

\frac{d}{d x} \arccos x = - \frac{1}{\sqrt{1 - x^{2}}} for - 1 < x < 1.

The minus sign is real: inverse cosine falls from $π$ at $x = - 1$ to 0 at $x = 1$ , so its slope is negative everywhere it exists.

Quick check · Match

No calculator. Tap a function, then its derivative.

4c. With the chain rule: an inside function

Most inverse trigonometric functions you meet have something other than $x$ inside. Call the inside $u$ and multiply by its derivative:

\frac{d}{d x} \arcsin u = \frac{1}{\sqrt{1 - u^{2}}} \cdot \frac{d u}{d x}, \frac{d}{d x} \arctan u = \frac{1}{1 + u^{2}} \cdot \frac{d u}{d x},

and $\frac{d}{d x} \arccos u = - \frac{1}{\sqrt{1 - u^{2}}} \cdot \frac{d u}{d x}$ .

For $y = \arctan (x^{3})$ , $u = x^{3}$ and $\frac{d u}{d x} = 3 x^{2}$ , so

\frac{d y}{d x} = \frac{1}{1 + (x^{3})^{2}} \cdot 3 x^{2} = \frac{3 x^{2}}{1 + x^{6}} .

Two slips are common. One drops $\frac{d u}{d x}$ and stops at $\frac{1}{1 + x^{6}}$ . The other squares only part of the inside: for $\arctan (5 x)$ the denominator is $1 + (5 x)^{2} = 1 + 25 x^{2}$ , not $1 + 5 x^{2}$ . Writing $u$ first, then squaring all of it, prevents both.

An inside function also moves the domain. $\arcsin (4 x)$ needs $- 1 \leq 4 x \leq 1$ , so it is defined only for $- \frac{1}{4} \leq x \leq \frac{1}{4}$ , and its derivative exists only for $- \frac{1}{4} < x < \frac{1}{4}$ .

Quick check · Calculate

No calculator. Let $f (x) = \arctan (3 x)$ . Find $f^{'} (1)$ . Give an exact decimal.

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Dropping the derivative of the inside, so that $\arctan (x^{3})$ gets the derivative $\frac{1}{1 + x^{6}}$ .	The rule $\frac{1}{1 + u^{2}}$ is the derivative with respect to $u$ . With respect to $x$ , the chain rule multiplies by $\frac{d u}{d x}$ .	Write $u$ and $\frac{d u}{d x}$ first, then the rule's structure, $\frac{1}{1 + u^{2}} \cdot \frac{d u}{d x}$ .
Squaring only part of the inside, writing $1 + 5 x^{2}$ for $\arctan (5 x)$ .	The rule squares $u$ , and here $u = 5 x$ , so $u^{2} = 25 x^{2}$ .	Put the whole inside in parentheses before squaring: $1 + (5 x)^{2}$ .
Giving inverse cosine the derivative $\frac{1}{\sqrt{1 - x^{2}}}$ , with no minus sign.	Inverse cosine falls from $π$ to 0 as $x$ runs from $- 1$ to 1, so its slope is negative.	Keep the minus sign: $\frac{d}{d x} \arccos x = - \frac{1}{\sqrt{1 - x^{2}}}$ .

Part 3 Watch it done Inverse sine of 2x, and where its derivative exists

Watch each step, then do the last one yourself.

Let $f (x) = \arcsin (2 x)$ . Find $f^{'} (x)$ , say where it exists, and evaluate it at $x = \frac{3}{10}$ .

Step 1 · Name the inside
$u = 2 x$ , so $\frac{d u}{d x} = 2$ .
Step 2 · Write the rule, then substitute
$f^{'} (x) = \frac{1}{\sqrt{1 - u^{2}}} \cdot \frac{d u}{d x} = \frac{1}{\sqrt{1 - (2 x)^{2}}} \cdot 2 = \frac{2}{\sqrt{1 - 4 x^{2}}}$ .
Step 3 · Say where it exists
$f$ needs $- 1 \leq 2 x \leq 1$ , so it is defined for $- \frac{1}{2} \leq x \leq \frac{1}{2}$ . Its derivative needs $1 - 4 x^{2} > 0$ , so it exists for $- \frac{1}{2} < x < \frac{1}{2}$ , and $\frac{3}{10}$ is inside that interval.
Why this step?
Why does $f^{'} (\frac{1}{2})$ fail to exist, even though $f (\frac{1}{2}) = \arcsin 1 = \frac{π}{2}$ is defined?
At $x = \frac{1}{2}$ the denominator $\sqrt{1 - 4 x^{2}}$ is 0, so $\frac{2}{\sqrt{1 - 4 x^{2}}}$ has no value. The graph of $f$ climbs ever more steeply as $x$ approaches $\frac{1}{2}$ , and no finite slope fits there.
Your step · Calculate

No calculator. Finish it. For $f (x) = \arcsin (2 x)$ , $f^{'} (x) = \frac{2}{\sqrt{1 - 4 x^{2}}}$ for $- \frac{1}{2} < x < \frac{1}{2}$ . What is $f^{'} (\frac{3}{10})$ ? Give an exact decimal.

Your answer

Part 4 Try it A product, and the sign of a derivative

Try it · Multiple choice

No calculator. Let $g (x) = x \arctan x$ . What is $g^{'} (1)$ ?

Try it · Sort it

No calculator. Tap a function, then the sign of its derivative wherever the derivative exists.

Derivative positive

Derivative negative

Part 5 Explain it Write it the way the exam asks

Free response on the exam is handwritten, so write your answer on paper first. Nothing you type here leaves this page. Then open the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. A drone lifts off from level ground 40 meters from an observer and rises straight up at a constant 10 meters per second. Let $θ (t)$ be the angle of elevation, in radians, from the observer to the drone $t$ seconds after liftoff, for $0 \leq t \leq 12$ .

(a) Explain why $θ (t) = \arctan (\frac{t}{4})$ .

(b) Find $θ^{'} (4)$ . Show the work that leads to your answer. Using correct units, interpret the meaning of this value in the context of the problem.

(c) The drone rises at a constant speed. Using $θ^{'} (t)$ , explain why the angle of elevation increases more and more slowly as $t$ increases.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) After $t$ seconds the drone is $10 t$ meters up, directly above a point 40 meters from the observer, so $\tan θ (t) = \frac{10 t}{40} = \frac{t}{4}$ . An angle of elevation is between 0 and $\frac{π}{2}$ , which is inside the interval $- \frac{π}{2} < θ < \frac{π}{2}$ where inverse tangent takes its values, so $θ (t) = \arctan (\frac{t}{4})$ .

(b) With $u = \frac{t}{4}$ , $θ^{'} (t) = \frac{1}{1 + u^{2}} \cdot \frac{d u}{d t} = \frac{1}{1 + {(\frac{t}{4})}^{2}} \cdot \frac{1}{4} = \frac{4}{16 + t^{2}}$ . So $θ^{'} (4) = \frac{4}{16 + 16} = \frac{1}{8}$ radian per second. Four seconds after liftoff, the angle of elevation from the observer to the drone is increasing at $\frac{1}{8}$ radian per second.

(c) $θ^{'} (t) = \frac{4}{16 + t^{2}}$ is positive, so the angle increases. As $t$ increases from 0 to 12, the denominator $16 + t^{2}$ increases, so $θ^{'} (t)$ decreases, from $\frac{1}{4}$ to $\frac{1}{40}$ radian per second. The angle keeps increasing, but at a smaller rate at each later time.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes $\tan θ (t) = \frac{10 t}{40} = \frac{t}{4}$ from the right triangle.
Says the angle lies between 0 and $\frac{π}{2}$ , inside the values inverse tangent takes, so $θ (t) = \arctan (\frac{t}{4})$ .
Uses the chain rule with the factor $\frac{1}{4}$ to get $θ^{'} (t) = \frac{4}{16 + t^{2}}$ , or an equal form, and $θ^{'} (4) = \frac{1}{8}$ .
Says the angle of elevation is increasing at $\frac{1}{8}$ radian per second at $t = 4$ , with units.
Explains that $θ^{'} (t)$ is positive and that its denominator $16 + t^{2}$ grows as $t$ increases, so the rate decreases.

Part 6 Check yourself An inside function given by a table

The two case questions use the same table. Answer them from it.

The function $f$ is differentiable for all real $x$ . Some of its values and the values of its derivative are in the table. Let $h (x) = \arctan (f (x))$ , and let $k (x) = \arcsin (f (x))$ wherever $- 1 \leq f (x) \leq 1$ .

Values of f and its derivative
$x$	$f (x)$	$f^{'} (x)$
1	$\frac{1}{2}$	3
2	3	5
3	$- 2$	4

Case question 1 · Calculate

No calculator. Find $h^{'} (2)$ . Give an exact decimal.

Your answer

Case question 2 · Multiple choice

No calculator. What is $k^{'} (1)$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the derivative of $y = \arccos (x^{2})$ for $- 1 < x < 1$ . Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. This time find the derivative of $y = \arctan x$ from the definition of an inverse and the chain rule. Tap a blank, then tap an answer.

Start from the definition: $y = \arctan x$ means , with $- \frac{π}{2} < y < \frac{π}{2}$ . Differentiating both sides with respect to $x$ gives $\sec^{2} y \cdot \frac{d y}{d x} = 1$ . The identity $\sec^{2} y = 1 + \tan^{2} y$ turns $\sec^{2} y$ into , so $\frac{d y}{d x}$ is .

Part 7 Keep it Bring back earlier ideas

This lesson stands on the derivative of an inverse function and on the derivative of tangent. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. The function $f$ is differentiable and one-to-one, with $f (2) = 5$ , $f^{'} (2) = 4$ , $f (5) = 9$ and $f^{'} (5) = 3$ . Let $g$ be the inverse of $f$ . What is $g^{'} (5)$ ?

From 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions · Calculate

No calculator. Let $f (x) = 3 \tan x - \cot x$ . Find $f^{'} (\frac{π}{4})$ .

Your answer

In three lines

Each inverse trigonometric function undoes its trigonometric function on one restricted interval, and that interval fixes the sign of the square root.
The definition of an inverse with the chain rule, or the inverse function formula, gives $\frac{1}{\sqrt{1 - x^{2}}}$ , $- \frac{1}{\sqrt{1 - x^{2}}}$ and $\frac{1}{1 + x^{2}}$ .
With an inside function, multiply by its derivative and square the whole inside, and check where the derivative exists.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When an inverse trigonometric function has something inside, write $u$ and $\frac{d u}{d x}$ first, then the rule's structure, such as $\frac{1}{1 + u^{2}} \cdot \frac{d u}{d x}$ , before any numbers go in. Square the whole inside, and check that the input you evaluate at lies in the open interval where the derivative exists.

Questions students ask

\sin^{- 1} x

the same as

\arcsin x

Yes. Both name inverse sine. The $- 1$ is not an exponent, so neither one means $\frac{1}{\sin x}$ .

Should I memorize these derivatives or work them out each time?

Know the three results well enough to use them at once. The derivation is your backup when you doubt a sign or a root: write $\sin y = x$ (or $\tan y = x$ , or $\cos y = x$ ), differentiate both sides, and replace the trigonometric function of $y$ with an expression in $x$ .

Why does the derivative of

\arcsin x

not exist at

x = 1

The formula $\frac{1}{\sqrt{1 - x^{2}}}$ divides by 0 there. On the graph, inverse sine gets steeper and steeper as $x$ approaches 1, and sine itself has slope 0 at $\frac{π}{2}$ , whose reciprocal does not exist.

Why are the derivatives of

\arcsin x

and

\arccos x

opposites?

For $- 1 \leq x \leq 1$ , $\arcsin x + \arccos x = \frac{π}{2}$ , a constant. A constant has derivative 0, so the two derivatives add to 0.

What about inverse cotangent, inverse secant and inverse cosecant?

The same two methods find their derivatives. Each one depends on the interval its trigonometric function is restricted to, so write the definition with its interval before you differentiate.

Course alignment, for teachers

AP Calculus AB topic 3.4, Unit 3: Differentiation: Composite, Implicit, and Inverse Functions.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.