NSWMaths Extension 1Syllabus dot point

What are the derivatives and antiderivatives of the inverse trigonometric functions?

Differentiate inverse trigonometric functions and integrate functions that involve them

A focused answer to the HSC Maths Extension 1 dot point on the calculus of inverse trig functions. The derivatives of $\arcsin$ , $\arccos$ and $\arctan$ , their chain-rule extensions, and integrals leading back to inverse trig, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to know the standard derivatives of $\arcsin$ , $\arccos$ and $\arctan$ , apply the chain rule to compositions, and integrate functions that produce these inverse-trig antiderivatives.

The answer

Standard derivatives

\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}, \qquad x \in (-1, 1).

\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1 - x^2}}, \qquad x \in (-1, 1).

\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}, \qquad x \in \mathbb{R}.

Note that $\arcsin$ and $\arccos$ have the same magnitude derivative but opposite signs, consistent with $\arcsin x + \arccos x = \frac{\pi}{2}$ .

Derivation (sketch)

Let $y = \arcsin x$ , so $\sin y = x$ . Differentiate implicitly:

$\cos y \cdot \frac{dy}{dx} = 1$ , so $\frac{dy}{dx} = \frac{1}{\cos y}$ .

For $y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ , $\cos y \ge 0$ , and $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ .

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

Similar reasoning yields the other two.

Chain rule extensions

\frac{d}{dx}\bigl(\arcsin f(x)\bigr) = \frac{f'(x)}{\sqrt{1 - [f(x)]^2}},

\frac{d}{dx}\bigl(\arctan f(x)\bigr) = \frac{f'(x)}{1 + [f(x)]^2}.

Standard antiderivatives

These mirror the derivatives:

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

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

Generalised forms:

\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin \frac{x}{a} + C, \qquad a > 0,

\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan \frac{x}{a} + C, \qquad a > 0.

Refer to the dedicated integration-of-inverse-trig dot point for completing-the-square techniques.

Worked example

Direct derivative

$\frac{d}{dx}(\arcsin(2 x)) = \frac{2}{\sqrt{1 - 4 x^2}}$ by the chain rule.

Chain rule on IMATH_1

$\frac{d}{dx}\bigl(\arctan(x^2)\bigr) = \frac{2 x}{1 + x^4}$ .

Product rule combination

$\frac{d}{dx}\bigl(x \arctan x\bigr) = \arctan x + x \cdot \frac{1}{1 + x^2} = \arctan x + \frac{x}{1 + x^2}$ .

Implicit derivative

Find $\frac{dy}{dx}$ if $y = \arcsin\!\left( \frac{x}{2} \right)$ at $x = 1$ .

$\frac{dy}{dx} = \frac{1/2}{\sqrt{1 - x^2/4}}$ at $x = 1$ gives $\frac{1/2}{\sqrt{3/4}} = \frac{1}{\sqrt{3}}$ .

Integral

Evaluate $\int_0^{1/2} \frac{1}{\sqrt{1 - 4 x^2}} \, dx$ .

$\sqrt{1 - 4 x^2} = \sqrt{1 - (2 x)^2}$ . Let $u = 2 x$ , $du = 2 dx$ .

$\int \frac{1}{\sqrt{1 - u^2}} \cdot \frac{1}{2} \, du = \frac{1}{2} \arcsin u + C = \frac{1}{2} \arcsin(2 x) + C$ .

Evaluate: $\frac{1}{2} \arcsin 1 - \frac{1}{2} \arcsin 0 = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$ .

Arctan integral

Evaluate $\int_0^{\sqrt{3}} \frac{1}{4 + x^2} \, dx$ .

$\frac{1}{4 + x^2}$ matches $\frac{1}{a^2 + x^2}$ with $a = 2$ .

$\int \frac{dx}{4 + x^2} = \frac{1}{2} \arctan \frac{x}{2} + C$ .

Evaluate: $\frac{1}{2} \arctan \frac{\sqrt{3}}{2} - 0$ . We need $\arctan \frac{\sqrt{3}}{2}$ . This is not a standard angle, so leave as $\frac{1}{2} \arctan \frac{\sqrt{3}}{2}$ .

(If the upper bound were $2$ instead of $\sqrt{3}$ , $\arctan 1 = \frac{\pi}{4}$ , giving $\frac{\pi}{8}$ .)