Derivatives of inverse trig functions — all six: Question bank

True or false — justify

Every symbol here means: $y = arcsin x$ etc., so $y$ is an angle and $x$ is a ratio the trig function of that angle produces.

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

False.

arccos

decreases — a bigger ratio

x

maps to a smaller angle — so its slope must be negative:

- \frac{1}{1 - x ^{2}}

arcsin x

and

arccos x

have the exact same derivative up to sign

True. Both share

\frac{1}{1 - x ^{2}}

;

arccos

just carries the minus. That is why

arcsin x + arccos x

has zero derivative and is constant

(\frac{π}{2})

The derivative of

arctan x

is defined for every real

x

True. Its denominator

1 + x^{2}

is never zero, so

\frac{1}{1 + x ^{2}}

exists for all

x

— unlike

arcsin

's, which dies at

x = \pm 1

\frac{d}{d x} arcsec x = \frac{1}{x x ^{2} - 1}

False. The

∣ x ∣

is mandatory.

arcsec

increases on both branches (

x < - 1

and

x > 1

), so its derivative must be positive everywhere; only

∣ x ∣

(always

\geq 0

) enforces that.

\frac{d}{d x} arcsin x

is largest (steepest) near

x = 0

False. At

x = 0

the denominator

1 - x^{2} = 1

, giving slope

1

. As

x \to \pm 1

the denominator

\to 0

and the slope blows up to

+ \infty

— the graph is steepest at the edges, flattest in the middle.

arctan x

and

arccot x

have graphs that are reflections, so their derivatives are opposite in sign

True.

arccot

decreases while

arctan

increases; their derivatives are

\pm \frac{1}{1 + x ^{2}}

, exact negatives of each other.

Because

sin y = x

, we may always write

cos y = 1 - x^{2}

with the positive root

True — but only for $arcsin$ . Its range

[- \frac{π}{2}, \frac{π}{2}]

keeps

cos y \geq 0

. For a different inverse with a different range you must recheck the sign; the positive root is earned, not automatic.

Spot the error

Each line states a "solution" someone wrote. Find and fix the mistake.

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

Missing chain-rule factor. The formula

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

applies to

arctan (u)

; here

u = 3 x

so you still owe

u^{'} = 3

. Correct:

\frac{3}{1 + 9 x ^{2}}

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

Two errors. The inside is

u = x^{2}

, so

u^{2} = x^{4}

(not

x^{2}

) under the root, and you must multiply by

u^{'} = 2 x

. Correct:

\frac{2 x}{1 - x ^{4}}

"Since

sec^{2} y = 1 + tan^{2} y

, differentiating

tan y = x

needs a square root at the end."

No root needed.

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

. The

sec^{2} y

is already

1 + x^{2}

— that clean cancellation is exactly why

arctan

's derivative has no square root.

\frac{d}{d x} arcsin x

x = 2

equals

\frac{1}{1 - 4} = \frac{1}{- 3}

Domain violation.

arcsin

only exists for

∣ x ∣ \leq 1

; there is no angle whose sine is

2

. The question itself is meaningless, not just the arithmetic.

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

, and I differentiated

cos y = x

to get

- sin y \frac{d y}{d x} = 1

, so

sin y = - 1 - x^{2}

Wrong sign on the root.

arccos

has range

[0, π]

, where

sin y \geq 0

, so

sin y = + 1 - x^{2}

. Then

\frac{d y}{d x} = \frac{- 1}{sin y} = \frac{- 1}{1 - x ^{2}}

— the minus comes from the algebra, not from the root.

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

by the power rule on

(1 + x^{2})^{- 1}

Power rule does not apply to

1/ (1 + x^{2})

because of the inner

x^{2}

. Read it backwards instead: it's exactly

\frac{d}{d x} arctan x

, so the integral is

arctan x + C

. See Integration by Inverse Trig.

\frac{d}{d x} arcsec x

x = \frac{1}{2}

\frac{1}{\frac{1}{2} \frac{1}{4} - 1}

Domain violation.

arcsec

needs

∣ x ∣ \geq 1

; there's no angle whose secant is

\frac{1}{2}

(secant never lands strictly between

- 1

and

1

). The expression under the root is negative for a reason.

Why questions

Why do the derivations start by writing

sin y = x

instead of differentiating

arcsin x

directly?

Because we can't differentiate

arcsin

from a known rule, but we can differentiate

sin

. Rewriting as

sin y = x

turns an unknown into a known relation we can implicitly differentiate and solve. See Implicit Differentiation.

Why does

arctan

get the tidy

\frac{1}{1 + x ^{2}}

while

arcsin

gets an ugly square root?

Because the identity that converts back to

x

differs.

arcsin

uses

cos y = 1 - sin^{2} y

(a root), while

arctan

uses

sec^{2} y = 1 + tan^{2} y

— a polynomial, no root — so its derivative stays rational.

Why must every "co-" function's derivative be negative?

Each co-function (

arccos

arccot

arccsc

) is decreasing on its range: as the ratio grows, the angle shrinks. A decreasing function has negative slope everywhere it's differentiable.

Why does the

∣ x ∣

appear only in

arcsec

and

arccsc

, never in the other four?

Because

tan y = \pm sec^{2} y - 1

carries a sign that flips between the two secant branches, yet

arcsec

increases on both. The

∣ x ∣

absorbs that sign flip so the derivative stays positive on both branches. The other four live on a single monotone piece, so no absolute value is needed.

Why does the slope of

arcsin x

explode to infinity as

x \to \pm 1

Geometrically, near a ratio of

\pm 1

a tiny change in the ratio corresponds to a huge change in the angle (the triangle nearly degenerates). Algebraically, the adjacent side

1 - x^{2} \to 0

, so

\frac{1}{1 - x ^{2}} \to \infty

Why do we need the range restriction of the inverse function at all?

Because

sin

(and every trig function) repeats, so "

sin y = x

" has infinitely many solutions

y

. Restricting the range picks one branch, which then pins down the sign of

cos y

sin y

, or

tan y

— and hence the sign of the derivative. See Inverse Functions and their Domains.

Why does knowing these six derivatives instantly give a family of integrals?

A derivative rule read right-to-left is an antiderivative. If

\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}

, then

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

by definition. Each formula donates its own integral.

Edge cases

What is

\frac{d}{d x} arcsin x

exactly at

x = \pm 1

Undefined (vertical tangent). The denominator

1 - x^{2} = 0

there, so the slope is infinite.

arcsin

is continuous at

\pm 1

but not differentiable — its graph has a vertical tangent.

\frac{d}{d x} arctan x

ever zero or negative?

Never.

\frac{1}{1 + x ^{2}}

is strictly positive for all real

x

; it approaches

0

x \to \pm \infty

but never reaches it. So

arctan

is strictly increasing but flattens at the tails.

x = 0

, do

arcsin

and

arctan

have the same slope?

Yes, both equal $1$ .

\frac{1}{1 - 0} = 1

and

\frac{1}{1 + 0} = 1

. Near the origin both curves hug the line

y = x

; they only diverge as

∣ x ∣

grows.

What happens to

\frac{d}{d x} arcsec x

∣ x ∣ \to \infty

\to 0^{+}

. Both

∣ x ∣

and

x^{2} - 1

grow without bound, so

\frac{1}{∣ x ∣ x ^{2} - 1} \to 0

—

arcsec

levels off toward its horizontal asymptotes.

Can you evaluate

\frac{d}{d x} arcsec x

x = 1

No.

x^{2} - 1 = 0

there, so the derivative is infinite (vertical tangent).

x = \pm 1

are the boundary points where the two branches begin.

arcsin x + arccos x = \frac{π}{2}

true for all valid

x

, and how does calculus prove it?

Yes, for all $x \in [- 1, 1]$ . Their derivatives are exact negatives, so the sum has derivative

0

and is constant. Evaluating at

x = 0

gives

0 + \frac{π}{2} = \frac{π}{2}

, fixing the constant.

What is the sign of

arccot

's derivative, and does that ever change across its domain?

Always negative, never changes.

- \frac{1}{1 + x ^{2}} < 0

for every real

x

, so

arccot

is strictly decreasing on its whole domain with no sign flip.

Recall One-line summary of every trap

Sign traps come from monotonicity (co-functions decrease), root-sign traps from range restrictions, $∣ x ∣$ traps from two-branch functions, chain-rule traps from forgotten inner derivatives, and domain traps from plugging in ratios the function can't produce. Name the category and the fix follows.

Connections

Derivatives of inverse trig functions — all six — the parent this bank drills
Implicit Differentiation — why every derivation starts with $sin y = x$
Pythagorean Identities — the source of every sign decision
Inverse Functions and their Domains — range restrictions that pin the signs
Chain Rule — the forgotten inner-derivative traps
Integration by Inverse Trig — reading the rules backwards
Derivatives of Trig Functions — the forward maps we invert