Single-Variable-Calculus · Unit 5 · Video 5 · Interactive Practice

Derivatives of the Arctangent and Arcsine

IKey Formulas

Formula	Name	Where it comes from
$\dfrac{d}{dy}\tan y = \dfrac{1}{\cos^2 y} = \sec^2 y$	Derivative of the tangent	Quotient rule on $\dfrac{\sin y}{\cos y}$ , then $\sin^2 y + \cos^2 y = 1$
$\tan y = x \ \Longrightarrow\ \dfrac{1}{\cos^2 y}\,y' = 1 \ \Longrightarrow\ y' = \cos^2 y$	Implicit differentiation	Chain rule applied to the defining equation
$\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^2}$	Arctangent derivative	Triangle: $\cos y = \dfrac{1}{\sqrt{1 + x^2}}$ ; valid for every $x$
$\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1 - x^2}}$	Arcsine derivative	$\cos y = +\sqrt{1 - x^2}$ on $-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}$ ; valid for $-1 < x < 1$

Key Insight: Implicit differentiation always answers in terms of $y$ . A right triangle (arctangent) or the identity $\sin^2 y + \cos^2 y = 1$ (arcsine) translates that answer back into $x$ , and the range of $y$ settles the sign — which is why neither formula contains a trigonometric function.

IIVisualization 1 — The Triangle That Removes the Trigonometry

The triangle behind $\cos^2(\arctan x)$ : opposite $x$ , adjacent $1$ , hypotenuse $\sqrt{1 + x^2}$ .

x 1.40

💡 The arctangent only ever needs $\cos^2 y$ , so the sign of $\cos y$ cannot affect it; the arcsine needs $\cos y$ itself, and there the sign is the whole question.

IIIVisualization 2 — One Height, Two Angles, One Branch

A height $x$ meets the circle twice, and the branch of $y$ decides which point counts.

height x = sin y 0.700

IVVisualization 3 — The Formula Is the Slope

Each formula is a slope you can see: gentle for $\arctan x$ , vertical for $\arcsin x$ near $\pm 1$ .

Curve x 1.000 Your slope 0.80

VQuiz Questions

Problem 1 · Evaluate the Arctangent Derivative

Given: $f(x) = \arctan x$ — find $f'(2)$ .

✅ Correct!

f'(x) = \dfrac{1}{1 + x^2}

, so

f'(2) = \dfrac{1}{1 + 4} = \dfrac{1}{5}

❌ Not quite. The denominator is

1 + x^2

, not

1 + x

. The square arrives from the hypotenuse

\sqrt{1 + x^2}

, which gets squared when

\cos y

becomes

\cos^2 y

❌ Close, but that is $\cos y$ , not $\cos^2 y$ . The triangle gives

\cos y = \dfrac{1}{\sqrt{1 + x^2}} = \dfrac{1}{\sqrt 5}

, and the derivative is its square.

❌ That is the tangent's derivative, not the arctangent's.

\dfrac{d}{dy}\tan y = \sec^2 y = 1 + x^2 = 5

; implicit differentiation then inverts it.

❌ Not quite. Differentiate

\tan y = x

to get

y' = \cos^2 y

, then rewrite

\cos^2 y

in terms of

x

Show solution

Write $y = \arctan x$ , which means $\tan y = x$ with $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ . Differentiate both sides with respect to $x$ , using the chain rule on the left:

\frac{1}{\cos^2 y}\,y' = 1 \qquad \Longrightarrow \qquad y' = \cos^2 y

Now rewrite $\cos^2 y$ in terms of $x$ with the right triangle: the side opposite $y$ has length $x$ , the adjacent side has length $1$ , so the hypotenuse is $\sqrt{1 + x^2}$ and

\cos y = \frac{1}{\sqrt{1 + x^2}}, \qquad \cos^2 y = \frac{1}{1 + x^2}

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

At $x = 2$ : $\ f'(2) = \dfrac{1}{1 + 2^2} = \dfrac{1}{5}$ .

Problem 2 · Why the Positive Root

Given: $y = \arcsin x$ , so $\sin y = x$ and implicit differentiation gives $y' = \dfrac{1}{\cos y}$ . The identity $\sin^2 y + \cos^2 y = 1$ only gives $\cos y = \pm\sqrt{1 - x^2}$ — which fact forces the $+$ sign?

✅ Correct! On that branch the angle

y

sweeps the right half of the circle, where the horizontal coordinate

\cos y

is never negative — so

\cos y = +\sqrt{1 - x^2}

and

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

❌ True, but it settles a different question. That is why the root is a real number; it says nothing about which of the two roots,

+

-

, equals

\cos y

❌ That identity is false.

\sqrt{u^2} = |u|

, which is exactly why a sign has to be decided from outside the algebra — here, from the range of

y

❌ The sign of $x$ is not the issue. Take

x = -\tfrac{1}{2}

: then

y = -\tfrac{\pi}{6}

and

\cos y = \tfrac{\sqrt 3}{2} > 0

. The sign of

\cos y

comes from the range of

y

, not from the sign of

x

❌ Not quite. The equation

\sin y = x

alone does not determine

y

— a branch must be chosen, and that choice is what fixes the sign of

\cos y

Show solution

A horizontal line at height $x$ meets the unit circle at two points, with horizontal coordinates $+\sqrt{1 - x^2}$ and $-\sqrt{1 - x^2}$ . Both are angles whose sine is $x$ , so $\sin y = x$ does not determine $y$ until a branch is chosen.

The arcsine's branch is

-\frac{\pi}{2} \le y \le \frac{\pi}{2}

which sweeps the right half of the circle: there the height runs from $-1$ to $1$ exactly once, and the horizontal coordinate satisfies $\cos y \ge 0$ . Hence

\cos y = +\sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}, \qquad \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}

At $x = \pm 1$ the two intersection points merge, $\cos y = 0$ , and the formula would divide by zero — so it holds for $-1 < x < 1$ .

For the arctangent the sign never mattered, because the answer was $\cos^2 y$ . Here $\cos y$ appears to the first power, and the branch decides its sign.

Problem 3 · Arcsine Inside a Chain Rule

Given: $g(x) = \arcsin(2x)$ — find $g'(x)$ and the values of $x$ for which it is defined.

What is the derivative?

For which $x$ is it defined?

✅ Correct!

g'(x) = \dfrac{2}{\sqrt{1 - 4x^2}}

, and the root needs

1 - 4x^2 > 0

, i.e.

|x| < \tfrac{1}{2}

❌ The chain rule factor is missing. With

u = 2x

\ \dfrac{d}{dx}\arcsin u = \dfrac{1}{\sqrt{1 - u^2}}\cdot\dfrac{du}{dx}

, and

\dfrac{du}{dx} = 2

❌ Check the square.

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

, not

2x^2

— the factor

2

gets squared too.

❌ That is the arctangent's shape.

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

comes from

\tan y = u

; the arcsine gives a square root,

\dfrac{1}{\sqrt{1 - u^2}}

❌ Not quite. Differentiate the outside arcsine at

u = 2x

, then multiply by

\dfrac{du}{dx}

❌ That is the arcsine's own domain. The number fed to the arcsine here is

2x

, so the requirement is

-1 < 2x < 1

❌ Check the domain. The expression under the root,

1 - 4x^2

, must be strictly positive.

Show solution

Set $u = 2x$ , so $g(x) = \arcsin u$ . The chain rule gives

g'(x) = \frac{1}{\sqrt{1 - u^2}}\cdot\frac{du}{dx} = \frac{1}{\sqrt{1 - (2x)^2}}\cdot 2 = \frac{2}{\sqrt{1 - 4x^2}}

The arcsine accepts an input between $-1$ and $1$ , and the derivative additionally needs the root to be non-zero:

1 - 4x^2 > 0 \quad \Longleftrightarrow \quad x^2 < \tfrac{1}{4} \quad \Longleftrightarrow \quad -\tfrac{1}{2} < x < \tfrac{1}{2}

Check at $x = 0$ : $g'(0) = \dfrac{2}{\sqrt 1} = 2$ , which is twice the slope $1$ of $\arcsin x$ at the origin — exactly what compressing the graph horizontally by a factor of $2$ should do.

Problem 4 · The Same Method on a New Function

Given: $y = \arccos x$ , which means $\cos y = x$ with $0 \le y \le \pi$ — run the same three steps (differentiate implicitly, rewrite in $x$ , let the range of $y$ fix the sign) and find $\dfrac{d}{dx}\arccos x$ .

✅ Correct!

\dfrac{d}{dy}\cos y = -\sin y

, and on

0 \le y \le \pi

the sine is non-negative, so

\sin y = +\sqrt{1 - x^2}

and the derivative is

-\dfrac{1}{\sqrt{1 - x^2}}

❌ The minus sign is missing. Differentiating

\cos y = x

gives

-\sin y\,y' = 1

, so

y' = -\dfrac{1}{\sin y}

. The arccosine decreases, so its derivative must be negative.

❌ That denominator belongs to the arctangent. Here the identity

\sin^2 y + \cos^2 y = 1

produces

\sin y = \sqrt{1 - x^2}

, a square root — not

1 + x^2

❌ Check what sits under the root. From

\cos y = x

\ \sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}

. On the domain

-1 < x < 1

the quantity

x^2 - 1

is negative, so that root is not even real.

❌ Not quite. Differentiate

\cos y = x

implicitly, solve for

y'

, then replace

\sin y

using

\sin^2 y + \cos^2 y = 1

with the sign chosen by

0 \le y \le \pi

Show solution

Step 1 — differentiate $\cos y = x$ implicitly (chain rule on the left, since $y$ depends on $x$ ):

-\sin y\;y' = 1 \qquad \Longrightarrow \qquad y' = -\frac{1}{\sin y}

Step 2 — rewrite in terms of $x$ using $\sin^2 y + \cos^2 y = 1$ and $\cos y = x$ :

\sin y = \pm\sqrt{1 - \cos^2 y} = \pm\sqrt{1 - x^2}

Step 3 — let the range of $y$ fix the sign. The arccosine's branch is $0 \le y \le \pi$ , the upper half of the unit circle, where the height $\sin y$ is never negative. So the positive root is the right one:

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

Consistency check: $\arcsin x + \arccos x = \dfrac{\pi}{2}$ for every $x$ in $[-1, 1]$ . Differentiating that constant gives $0$ , so the two derivatives must be negatives of each other — and they are.

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