Why can't I just use 1/f'(b) for the inverse derivative?

Because the formula requires evaluating f' at f^-1(b), not at b. The point b is an output of f, but f' takes inputs. You must first find the input a with f(a)=b, then compute f'(a) and reciprocate. Only if b happened to equal f^-1(b) would the shortcut accidentally work.

Do I have to memorize the derivatives of arccot, arcsec, and arccsc too?

The three you must know cold are arcsin, arccos, and arctan, since they appear most often and drive the standard integration formulas. The other three are less common on the exam, but knowing that arccos, arccot, and arccsc derivatives carry a negative sign (mirroring their pairs) helps you reconstruct them if needed.

What happens if f'(f^-1(b))=0?

Then the inverse-function formula produces division by zero, so (f^-1)'(b) does not exist. Geometrically, a horizontal tangent on f reflects to a vertical tangent on f^-1, which has undefined slope. On a free-response question you should explicitly note this case.

How is this lesson connected to implicit differentiation?

The inverse trig derivatives are derived using implicit differentiation from U3.2. For example, from y=x you differentiate implicitly to get ^2 y· y'=1, then solve for y'. The inverse-function formula itself also comes from differentiating f(f^-1(x))=x with the Chain Rule.

AP-CALCBC-3.3-3.4

U3.3 Derivatives of Inverse Functions

Master the inverse-function derivative formula and the derivatives of arcsin, arccos, and arctan for AP Calculus BC, with worked examples and practice.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U3.3 Derivatives of Inverse Functions, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U3.3 Derivatives of Inverse Functions free

What this lesson covers

How fast does an inverse function change? You don't need a formula for

f^{-1}

to answer that — you just need one clever relationship. In this lesson you'll learn the inverse-function derivative formula

(f^{-1})'(b)=\dfrac{1}{f'(f^{-1}(b))}

and see why it falls straight out of the Chain Rule you studied in U3.1.

You'll also lock in the three inverse trig derivatives the AP exam expects you to know cold: arcsin, arccos, and arctan. These show up in multiple-choice questions, in later integration units, and inside chain-rule composites. By the end you'll be able to evaluate an inverse derivative at a point without ever solving for the inverse algebraically.

Why the inverse formula works

f

is a one-to-one differentiable function, its inverse

f^{-1}

undoes it:

f(f^{-1}(x))=x

. Differentiate both sides using the Chain Rule from U3.1. The derivative of the right side is

1

, and the left side gives

f'(f^{-1}(x))\cdot (f^{-1})'(x)

. Setting these equal and solving:

(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}.

Evaluated at a specific input

b

, this becomes

(f^{-1})'(b)=\dfrac{1}{f'(f^{-1}(b))}

.

The geometric meaning is clean. The graph of

f^{-1}

is the reflection of

f

across the line

y=x

. Reflection swaps the roles of rise and run, so the slope of

f^{-1}

at a point is the reciprocal of the slope of

f

at the mirror-image point. If

f

passes through

(a,b)

with slope

m

, then

f^{-1}

passes through

(b,a)

with slope

\frac{1}{m}

.

A common misconception is to write

(f^{-1})'(b)=\frac{1}{f'(b)}

. That is wrong. You must first find the input

a=f^{-1}(b)

— the

x

-value where

f(a)=b

— and evaluate

f'

there, not at

b

itself. The formula requires plugging

f^{-1}(b)

into

f'

, so identifying that mirror point is the key step.

How to apply it step by step

Almost every AP problem of this type gives you

f

(or a table of values) and asks for

(f^{-1})'(b)

at one number. Follow a fixed routine so you never mix up which value goes where.

Step	Action
1	Find $a$ such that $f(a)=b$ ; this is $a=f^{-1}(b)$
2	Compute the derivative $f'(x)$
3	Evaluate $f'(a)$
4	Take the reciprocal: $(f^{-1})'(b)=\dfrac{1}{f'(a)}$

Step 1 is where students slip. Because

f^{-1}

is often impossible to write explicitly, you solve

f(a)=b

by inspection, factoring, or reading a table. For example, if

f(x)=x^3+x

and

b=2

, test values:

f(1)=1+1=2

, so

a=1

.

The exam loves table problems because they force you to use the formula rather than algebra. A table gives

f(a)

and

f'(a)

for a few points; you locate the row where

f(a)=b

, read off

f'(a)

, and reciprocate. If

f'(a)=0

at the needed point, the inverse has a vertical tangent and

(f^{-1})'(b)

does not exist.

Derivatives of arcsin, arccos, and arctan

Inverse trig derivatives are really just the inverse-function formula applied to sine, cosine, and tangent, but you should memorize the results. Here are the three the AP exam tests most:

Function	Derivative	Domain notes
$\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$	$-1<x<1$
$\arccos x$	$-\dfrac{1}{\sqrt{1-x^2}}$	$-1<x<1$
$\arctan x$	$\dfrac{1}{1+x^2}$	all real $x$

Notice the pairing:

\arccos x

is just the negative of

\arcsin x

's derivative, and their sum

\arcsin x+\arccos x=\frac{\pi}{2}

is constant, which is why the derivatives cancel. The other three inverse trig functions —

\operatorname{arccot}

\operatorname{arcsec}

, and

\operatorname{arccsc}

— have their own formulas, but arcsin, arccos, and arctan are the essentials.

To derive

\arctan

: let

y=\arctan x

, so

\tan y=x

. Differentiate implicitly (U3.2):

\sec^2 y\cdot y'=1

, giving

y'=\frac{1}{\sec^2 y}=\frac{1}{1+\tan^2 y}=\frac{1}{1+x^2}

. The same implicit trick, using

\cos^2+\sin^2=1

, produces the arcsin formula.

Chain rule with inverse trig

On the exam these derivatives rarely appear bare; they are usually wrapped inside a composition, so combine them with the Chain Rule. If

u

is a function of

x

, then

\frac{d}{dx}\arcsin(u)=\frac{u'}{\sqrt{1-u^2}},\qquad \frac{d}{dx}\arctan(u)=\frac{u'}{1+u^2}.

For instance,

\frac{d}{dx}\arctan(3x)=\frac{3}{1+9x^2}

, and

\frac{d}{dx}\arcsin(x^2)=\frac{2x}{\sqrt{1-x^4}}

. The pattern is always the base formula with

x

replaced by the inside function, multiplied by the inside function's derivative.

A frequent error is forgetting to square the inside function inside the radical or denominator: for

\arcsin(u)

the denominator is

\sqrt{1-u^2}

, not

\sqrt{1-x^2}

. Another is dropping the negative sign on

\arccos

. Keep track of signs, because a sign error converts a correct method into a wrong answer that graders and multiple-choice distractors are designed to catch.

These derivatives also anticipate later integration units, where

\int \frac{dx}{1+x^2}=\arctan x+C

and

\int \frac{dx}{\sqrt{1-x^2}}=\arcsin x+C

. Recognizing the derivative forms now makes those antiderivatives automatic.

How the AP exam tests this topic

Expect this material in both multiple-choice and free-response settings. The most common multiple-choice item gives a specific function and asks for

(f^{-1})'

at a value, testing whether you evaluate

f'

f^{-1}(b)

rather than at

b

. A close second is a table-based question where you must locate the correct row.

Inverse trig derivatives appear as standalone differentiation problems and as pieces of larger chain-rule or product-rule expressions. Calculator-active questions may ask you to evaluate an inverse derivative numerically, while non-calculator questions test the exact formulas.

Test format	Typical task
Multiple choice	Evaluate $(f^{-1})'(b)$ from a rule or table
Multiple choice	Differentiate $\arctan(g(x))$ or similar composite
Free response	Justify existence of $(f^{-1})'$ using $f'\neq 0$

To justify that

f^{-1}

is differentiable at

b

, state that

f

is differentiable and one-to-one and that

f'(f^{-1}(b))\neq 0

. If

f'

equals zero there, the reciprocal is undefined and you should say the derivative does not exist. Clear reasoning earns justification points even when the arithmetic is short.

Key terms

Inverse function.: A function $f^{-1}$ that reverses $f$ , so that $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$ . It exists when $f$ is one-to-one.
Inverse-function derivative formula.: The rule $(f^{-1})'(b)=\dfrac{1}{f'(f^{-1}(b))}$ , giving the slope of an inverse without solving for it explicitly.
One-to-one.: A function that never repeats an output value, so it passes the horizontal line test and has an inverse.
arcsin.: The inverse sine function; its derivative is $\dfrac{1}{\sqrt{1-x^2}}$ on $-1<x<1$ .
arccos.: The inverse cosine function; its derivative is $-\dfrac{1}{\sqrt{1-x^2}}$ on $-1<x<1$ .
arctan.: The inverse tangent function; its derivative is $\dfrac{1}{1+x^2}$ for all real $x$ .
Reciprocal slope.: The geometric reason inverses behave as they do: reflecting across $y=x$ turns a slope $m$ into $\frac{1}{m}$ .

Worked example

Let

f(x)=x^3+2x+1

. Given that

f

is one-to-one, find

(f^{-1})'(4)

Use

(f^{-1})'(4)=\dfrac{1}{f'(f^{-1}(4))}

.

Step 1: Find

a=f^{-1}(4)

, the input with

f(a)=4

. Test

a=1

f(1)=1^3+2(1)+1=1+2+1=4

. So

a=1

.

Step 2: Differentiate:

f'(x)=3x^2+2

.

Step 3: Evaluate at

a=1

f'(1)=3(1)^2+2=3+2=5

.

Step 4: Take the reciprocal:

(f^{-1})'(4)=\dfrac{1}{5}

.

Notice we never found a formula for

f^{-1}

— solving

x^3+2x+1=y

for

x

would be miserable. The formula lets us bypass that entirely by working at the mirror point

x=1

Practice questions

g(x)=\arctan(x^2)

, what is

g'(x)

$\dfrac{1}{1+x^4}$
$\dfrac{2x}{1+x^4}$
$\dfrac{2x}{1+x^2}$
$\dfrac{1}{1+x^2}$

Answer: $\dfrac{2x}{1+x^4}$

Apply the chain rule with

u=x^2

. The arctan derivative gives

\frac{u'}{1+u^2}

. Here

u'=2x

and

u^2=(x^2)^2=x^4

, so

g'(x)=\frac{2x}{1+x^4}

. Choice with

1+x^2

forgets to square the inside function; the choice without

2x

forgets the chain rule factor.

The function

f

is differentiable and one-to-one. A table gives

f(2)=5

and

f'(2)=3

, and

f(5)=9

and

f'(5)=7

. Find

(f^{-1})'(5)

Answer: $(f^{-1})'(5)=\dfrac{1}{3}$

You need the input

a

with

f(a)=5

. From the table,

f(2)=5

, so

a=2

. Then

(f^{-1})'(5)=\frac{1}{f'(2)}=\frac{1}{3}

. The trap is using

f'(5)=7

; that would answer

(f^{-1})'(9)

, not

(f^{-1})'(5)

. Always evaluate

f'

f^{-1}(b)

, which is the row where the output equals

b

Explain why

\arcsin x + \arccos x

has a derivative of zero, and what that tells you about the two functions.

Answer: The derivatives are opposites, so their sum is constant.

\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}

and

\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}}

. Adding gives

0

. A zero derivative on an interval means the sum is constant. Evaluating at

x=0

gives

\arcsin 0+\arccos 0=0+\frac{\pi}{2}=\frac{\pi}{2}

, so

\arcsin x+\arccos x=\frac{\pi}{2}

throughout the domain.

FAQ

Why can't I just use $1/f'(b)$ for the inverse derivative?: Because the formula requires evaluating $f'$ at $f^{-1}(b)$ , not at $b$ . The point $b$ is an output of $f$ , but $f'$ takes inputs. You must first find the input $a$ with $f(a)=b$ , then compute $f'(a)$ and reciprocate. Only if $b$ happened to equal $f^{-1}(b)$ would the shortcut accidentally work.
Do I have to memorize the derivatives of arccot, arcsec, and arccsc too?: The three you must know cold are arcsin, arccos, and arctan, since they appear most often and drive the standard integration formulas. The other three are less common on the exam, but knowing that arccos, arccot, and arccsc derivatives carry a negative sign (mirroring their pairs) helps you reconstruct them if needed.
What happens if $f'(f^{-1}(b))=0$ ?: Then the inverse-function formula produces division by zero, so $(f^{-1})'(b)$ does not exist. Geometrically, a horizontal tangent on $f$ reflects to a vertical tangent on $f^{-1}$ , which has undefined slope. On a free-response question you should explicitly note this case.
How is this lesson connected to implicit differentiation?: The inverse trig derivatives are derived using implicit differentiation from U3.2. For example, from $\tan y=x$ you differentiate implicitly to get $\sec^2 y\cdot y'=1$ , then solve for $y'$ . The inverse-function formula itself also comes from differentiating $f(f^{-1}(x))=x$ with the Chain Rule.

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