Derivatives of inverse trig functions — all six

The master method (learn this ONCE, reuse six times)

WHY this works: the inverse equation $\sin y = x$ is an honest relation between $x$ and $y$ . We don't yet know $\frac{dy}{dx}$ , but we can differentiate the relation and solve for it. This is exactly the trick that turns "I can't differentiate $\arcsin$ directly" into "I can differentiate $\sin$ , no problem."

Deriving $\frac{d}{dx}\arcsin x$ from scratch

$\sin y = x$

Differentiate both sides ( $\frac{d}{dx}$ ), using the chain rule on the left: $\cos y \cdot \frac{dy}{dx} = 1$

Why this step? The left side is $\sin(\text{something})$ , so its derivative is $\cos(\text{something})\times(\text{derivative of something})$ , and that something is $y$ .

Solve: $\frac{dy}{dx} = \frac{1}{\cos y}$

Now we must write $\cos y$ in terms of $x$ , not $y$ (a derivative should be a function of $x$ ). Use $\sin^2 y + \cos^2 y = 1$ : $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$

Why the + root? Because $\arcsin$ has range $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ , where $\cos y \ge 0$ . So we take the positive square root.

$\boxed{\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^2}}, \qquad -1 < x < 1}$

Figure — Derivatives of inverse trig functions — all six

The other five — same recipe, different trig identity

Formula All six (the "20%" that gives 80%)

$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}$ $\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}$ $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$ $\frac{d}{dx}\operatorname{arccot} x = -\frac{1}{1+x^2}$ $\frac{d}{dx}\operatorname{arcsec} x = \frac{1}{|x|\sqrt{x^2-1}}$ $\frac{d}{dx}\operatorname{arccsc} x = -\frac{1}{|x|\sqrt{x^2-1}}$

PATTERN: the three "co-" functions are just the negatives of their partners. Memorize three, flip the sign for the other three.

$\arccos$ : $\cos y = x \Rightarrow -\sin y\,\frac{dy}{dx}=1 \Rightarrow \frac{dy}{dx}=\frac{-1}{\sin y}=\frac{-1}{\sqrt{1-x^2}}$ (range $[0,\pi]$ , so $\sin y\ge 0$ ).

$\arctan$ : $\tan y = x \Rightarrow \sec^2 y\,\frac{dy}{dx}=1 \Rightarrow \frac{dy}{dx}=\frac{1}{\sec^2 y}=\frac{1}{1+\tan^2 y}=\frac{1}{1+x^2}.$ Why $\sec^2 y = 1+\tan^2 y$ ? It's the Pythagorean identity divided by $\cos^2$ . No square root needed — that's why $\arctan$ 's derivative is so clean.

$\operatorname{arcsec}$ : $\sec y = x \Rightarrow \sec y\tan y\,\frac{dy}{dx}=1 \Rightarrow \frac{dy}{dx}=\frac{1}{\sec y\tan y}$ . Now $\tan y = \pm\sqrt{\sec^2y-1}=\pm\sqrt{x^2-1}$ and after handling the range carefully you get $\frac{1}{|x|\sqrt{x^2-1}}$ . The $|x|$ is what makes the derivative positive everywhere arcsec is increasing.

Worked examples

Common mistake Steel-manning the classic errors

Mistake A — forgetting the chain rule. Writing $\frac{d}{dx}\arctan(3x)=\frac{1}{1+9x^2}$ . Why it feels right: you "applied the formula." The fix: the formula is for $\arctan(u)$ where $u=x$ . When $u\ne x$ you owe an extra $u'$ . Answer: $\frac{3}{1+9x^2}$ .

Mistake B — wrong sign on the "co-" functions. Thinking $\frac{d}{dx}\arccos x = \frac{1}{\sqrt{1-x^2}}$ . Why it feels right: it looks like arcsin's. The fix: arccos decreases (bigger ratio → smaller angle), so its slope must be negative.

Mistake C — dropping the $|x|$ in arcsec. Writing $\frac{1}{x\sqrt{x^2-1}}$ . Why it feels right: it falls straight out of the algebra. The fix: arcsec is increasing on both branches, so its derivative is positive everywhere — the $|x|$ enforces this.

Mistake D — ignoring the domain. Computing $\frac{d}{dx}\arcsin x$ at $x=2$ . Impossible: $\arcsin$ only exists for $|x|\le 1$ , and $\sqrt{1-x^2}$ isn't real beyond that.

Recall Feynman: explain it to a 12-year-old

Imagine a ramp. "Sine" tells you: if I tilt the ramp this many degrees, how steep does it look? The inverse asks the reverse: I see this steepness — how much did I tilt it? Now we want to know how fast that answer changes when the steepness changes a tiny bit. Trick: we already know how sine behaves, so we flip the question around using a triangle, draw the triangle with sides $x$ , $1$ , and $\sqrt{1-x^2}$ , and read off the answer. Near a ratio of $\pm 1$ the angle changes super fast (that's why $\sqrt{1-x^2}\to 0$ blows the slope up!).

Flashcards

Derivative of

\arcsin x

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

Derivative of

\arccos x

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

Derivative of

\arctan x

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

Derivative of

\operatorname{arccot} x

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

Derivative of

\operatorname{arcsec} x

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

Derivative of

\operatorname{arccsc} x

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

First step to derive

\frac{d}{dx}\arcsin x

Write

\sin y = x

and implicitly differentiate

Why does arccos get a minus sign?

It is decreasing, so its slope is negative

Why

|x|

in arcsec's derivative?

arcsec increases on both branches, so derivative must stay positive

\frac{d}{dx}\arctan(3x)

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

(chain rule gives the extra 3)

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

\arctan x + C

What identity proves

\arcsin x+\arccos x=\pi/2

Their derivatives cancel to 0, so the sum is constant; check

x=0

Why

+

root for

\cos y

in arcsin derivation?

Range of arcsin is

[-\pi/2,\pi/2]

where

\cos y\ge 0

Connections

Implicit Differentiation — the engine behind every derivation here
Chain Rule — needed for composite inverse-trig functions
Pythagorean Identities — convert $\cos y, \sec y$ back into $x$
Inverse Functions and their Domains — range restrictions fix the signs
Integration by Inverse Trig — these formulas read backwards
Derivatives of Trig Functions — the forward versions we invert

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, inverse trig function ka matlab simple hai: $\arcsin x$ ka matlab hai "wo angle jiska sine $x$ hai." Ratio do, angle wapas lo. Ab inka derivative yaad karne ki zarurat nahi — hum har baar ek hi trick se nikaalenge. Trick yeh hai: $y=\arcsin x$ likho, isse $\sin y = x$ banta hai, aur dono taraf $x$ ke respect mein differentiate kar do (implicit differentiation). Chain rule se $\cos y \cdot y' = 1$ aata hai, phir $y' = 1/\cos y$ , aur Pythagoras se $\cos y = \sqrt{1-x^2}$ . Bas, ho gaya.

Ek chhota sa right-triangle banao — opposite $=x$ , hypotenuse $=1$ , toh adjacent automatically $\sqrt{1-x^2}$ aa jaata hai. Yeh triangle picture se aap formula bhool hi nahi sakte. Yaad rakho: jo "co-" wale functions hain (arccos, arccot, arccsc), unka derivative bas negative sign ke saath same hota hai — kyunki yeh decreasing functions hain.

Sabse common galti? Chain rule bhool jaana. $\arctan(3x)$ ka derivative $\frac{1}{1+9x^2}$ nahi, balki $\frac{3}{1+9x^2}$ hai — andar wale $3x$ ka derivative $3$ multiply karna padta hai. Doosri galti: arcsec mein $|x|$ chhod dena. Yeh modulus isliye hai kyunki arcsec dono side increasing hai, toh slope hamesha positive.

Yeh formulas kyun important? Kyunki integration mein ulta kaam karte hain. $\int \frac{dx}{1+x^2} = \arctan x + C$ — yeh seedha inverse trig derivative ko backward padhne se aata hai. Toh in chhe formulas ko theek se yaad karo, exam mein integral aur derivative dono jagah kaam aayenge. 80/20 rule: teen formula yaad karo, baaki teen ka sign flip karo!

Derivatives of inverse trig functions — all six

The master method (learn this ONCE, reuse six times)

Deriving $\frac{d}{dx}\arcsin x$ from scratch

The other five — same recipe, different trig identity

Worked examples

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus I — Limits & Derivatives

Connections