Derivatives of inverse trig functions — all six: Visual walkthrough

Before we start, one promise: we will not use a single symbol until it has a plain-word meaning and a place in a drawing. Let's meet the cast.

Step 1 — What "arcsin" even asks

WHAT. A machine called $sin$ eats an angle and spits out a ratio — a plain number. Its inverse, written $arcsin$ , runs the machine backwards: you hand it the ratio, it tells you which angle produced it.

WHY. We cannot differentiate $arcsin$ head-on — we have no rule for "the angle whose sine is $x$ ." But we do know the forward machine $sin$ inside out. So the whole strategy is: turn the backward question into a forward one we can already handle.

PICTURE. On the left, angle in → ratio out. On the right, the arrow reversed: ratio in → angle out.

Figure — Derivatives of inverse trig functions — all six

Step 2 — Flip the backward question into a forward one

WHAT. Apply the forward machine $sin$ to both sides of $y = arcsin x$ . Since $sin$ and $arcsin$ undo each other, the right side collapses: $sin y = x .$

WHY. Look what just happened. The equation $sin y = x$ contains only the forward function. It is an honest relationship tying $y$ and $x$ together — no inverse in sight. We have escaped the thing we could not differentiate.

PICTURE. The two machines cancel like a key and its lock; what survives is the clean relation $sin y = x$ .

Step 3 — Draw the relation as a right triangle

WHAT. The statement $sin y = x$ is really $sin y = \frac{x}{1}$ . Recall that for an angle in a right triangle, $sin y = \frac{opposite side}{hypotenuse} .$ So we can build a triangle that obeys our equation: put the opposite side $= x$ and the hypotenuse $= 1$ .

WHY. A triangle is a memory device. Once we place two sides, Pythagoras hands us the third for free — and that third side is exactly the quantity we'll need in a moment.

PICTURE. Angle $y$ at the corner, vertical (opposite) side labelled $x$ , slanted hypotenuse labelled $1$ , and the horizontal (adjacent) side still unknown — call it $a$ .

Now the cosine of $y$ is readable straight off the triangle: $cos y = \frac{adjacent}{hypotenuse} = \frac{1 - x ^{2}}{1} = 1 - x^{2} .$

Step 4 — Differentiate the honest relation

WHAT. Take $\frac{d}{d x}$ of both sides of $sin y = x$ . The right side is easy: the derivative of $x$ with respect to $x$ is $1$ . The left side needs the Chain Rule because $y$ is itself a function of $x$ (it changes as we slide $x$ ): $deriv of s i n cos y \cdot deriv of the inside \frac{d y}{d x} = 1.$

WHY. This is Implicit Differentiation: we never solved for $y$ explicitly, yet by differentiating the whole relation we can solve for the slope $\frac{d y}{d x}$ afterwards. The tool exists precisely because we could not isolate $arcsin$ cleanly.

PICTURE. A tiny nudge $d x$ in the ratio produces a tiny nudge $d y$ in the angle; the factor $cos y$ is the "gearing" between them that the chain rule exposes.

Step 5 — Solve for the slope, then translate back to $x$

WHAT. Divide both sides by $cos y$ : $\frac{d y}{d x} = \frac{1}{c o s y} .$ But a derivative should be a function of $x$ , not of the hidden angle $y$ . So we swap $cos y$ for its triangle value from Step 3, $cos y = 1 - x^{2}$ : $\frac{d}{d x} arcsin x = \frac{1}{1 - x ^{2}}$

WHY. The triangle is the bridge. It converts the angle-language answer $\frac{1}{c o s y}$ into the ratio-language answer $\frac{1}{1 - x ^{2}}$ that only mentions $x$ .

PICTURE. The slope of the $arcsin$ curve at ratio $x$ equals one over the adjacent side of our triangle — read the answer straight off the drawing.

Step 6 — Why the positive root, and the domain $- 1 < x < 1$

WHAT. In Step 3 we took $cos y = + 1 - x^{2}$ , throwing away the negative option. We must justify that, and also say where the formula is even allowed to live.

WHY the plus sign. The angle $y = arcsin x$ is forced to lie in $[- \frac{π}{2}, \frac{π}{2}]$ (Step 1's restriction). Across that entire band, cosine is never negative — it runs from $0$ up to $1$ and back to $0$ . So $cos y \geq 0$ , and the positive root is the only correct choice.

WHY the open interval. Watch what happens at the ends:

At $x = \pm 1$ : $1 - x^{2} = 0 = 0$ , so $\frac{1}{1 - x ^{2}}$ blows up — the tangent to the $arcsin$ curve becomes vertical. The derivative simply does not exist there.
Beyond $∣ x ∣ > 1$ : $1 - x^{2} < 0$ , so the square root is not a real number, and neither is $arcsin x$ itself. There is no angle whose sine exceeds $1$ .

PICTURE. The cosine graph over the arcsin band sits on or above the axis (never dips negative), while the derivative's spikes shoot to infinity exactly at $x = \pm 1$ .

Step 7 — Sanity check the shape near the edges

WHAT. Test the formula's behaviour, not just its algebra. As $x \to 0$ , the slope is $\frac{1}{1 - 0} = 1$ . As $x \to \pm 1$ , the slope $\to + \infty$ .

WHY. Near a ratio of $0$ the angle changes at the same pace as the ratio (slope $1$ ). Near a ratio of $\pm 1$ the sine curve is nearly flat, so a tiny change in ratio demands a huge swing in angle — hence the exploding slope. The formula predicts exactly this, which is strong evidence it's right.

PICTURE. The $arcsin$ curve is gentle in the middle (slope $\approx 1$ ) and rears up almost vertically as it approaches $x = \pm 1$ .

The one-picture summary

Everything above, compressed: the backward question becomes $sin y = x$ , becomes a triangle, becomes a slope read off the adjacent side.

Recall Feynman retelling — the whole movie in plain words

We wanted to know how fast the angle answer changes when the ratio we feed in wiggles a little. We couldn't attack that directly, so we flipped it: sine of our angle just equals the ratio, $sin y = x$ . That sentence is a triangle — opposite side $x$ , hypotenuse $1$ — and Pythagoras fills the last side with $1 - x^{2}$ . Nudging both sides of $sin y = x$ gives $cos y \cdot (angle wiggle) = (ratio wiggle)$ , so the angle wiggle is one over $cos y$ — and $cos y$ is just the adjacent side $1 - x^{2}$ . Answer: $\frac{1}{1 - x ^{2}}$ . It's positive because our angle lives where cosine never goes negative; it explodes at $x = \pm 1$ because there the sine curve flattens and the angle has to race; and it's meaningless past $x = \pm 1$ because no angle has a sine bigger than one.

Recall

First move to differentiate $arcsin x$ ::: write $sin y = x$ and differentiate implicitly Where does $1 - x^{2}$ come from ::: the adjacent side of the triangle with opposite $x$ , hypotenuse $1$ Why the positive root for $cos y$ ::: arcsin's angle lives in $[- π /2, π /2]$ where $cos y \geq 0$ What happens to the slope at $x = \pm 1$ ::: it blows up to $+ \infty$ ; the tangent goes vertical Value of the slope at $x = \frac{1}{2}$ ::: $\frac{2}{3}$

Connections

Inverse Functions and their Domains — the range $[- \frac{π}{2}, \frac{π}{2}]$ that fixes the sign
Implicit Differentiation — the engine that turns $sin y = x$ into a slope
Chain Rule — why the left side gets a trailing $\frac{d y}{d x}$
Pythagorean Identities — supplies the adjacent side $1 - x^{2}$
Derivatives of Trig Functions — the forward $\frac{d}{d x} sin y = cos y$ we invert
Integration by Inverse Trig — this derivative read backwards