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Module 8 · L10 of 20 ~35 min ⚡ +95 XP available

Derivative of $\sin^{-1} x$

One formula, $\dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1-x^2}}$, is the gateway to integrating $\dfrac{1}{\sqrt{a^2-x^2}}$ — one of the most common HSC integration targets. In this lesson you'll derive it cleanly from the inverse function rule, understand why the domain is $(-1,1)$, and build fluency with the chain-rule extension for compound arguments like $\sin^{-1}(3x)$.

Today's hook — The curve $y = \sin^{-1}x$ has vertical tangents at its endpoints $x = \pm 1$. That means the derivative blows up there. What do you think happens to $\dfrac{1}{\sqrt{1-x^2}}$ as $x \to 1^-$? Jot your prediction before card 05.

0/5QUESTS

Recall — your gut answer first

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From Lesson 9 you know that $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$ when $y = f^{-1}(x)$. Before deriving it — predict what $\dfrac{d}{dx}(\sin^{-1}x)$ should look like: will the denominator involve a square root? A trig function? Both? Write your reasoning.

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The big idea: implicit differentiation on $x = \sin y$

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Three-step strategy: rewrite, differentiate implicitly, use the Pythagorean identity.

Starting from $y = \sin^{-1}x$:

Step 1 — Rewrite: $x = \sin y$, with $y \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$.

Step 2 — Differentiate: $1 = \cos y \cdot \dfrac{dy}{dx}$, so $\dfrac{dy}{dx} = \dfrac{1}{\cos y}$.

Step 3 — Substitute: $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1-x^2}$ (positive on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$).

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

Domain: $(-1, 1)$

The derivative $\dfrac{1}{\sqrt{1-x^2}}$ is only defined for $-1 < x < 1$. At $x = \pm1$ the denominator is zero and the tangent is vertical.

Always positive

$\cos y \geq 0$ on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, so $\dfrac{dy}{dx} > 0$ for all $x \in (-1,1)$. The $\sin^{-1}$ curve is always increasing.

Chain rule extension

$\dfrac{d}{dx}[\sin^{-1}(u)] = \dfrac{u'}{\sqrt{1-u^2}}$ whenever $u = g(x)$. Lesson 12 covers this in full.

What you'll master

Know

Key facts

$\dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1-x^2}}$, domain $(-1,1)$
The restricted domain of $\sin^{-1}$ is $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$
The derivative is undefined at $x = \pm 1$ (vertical tangent)

Understand

Concepts

How implicit differentiation of $x = \sin y$ delivers the result
Why $\cos y = \sqrt{1-x^2}$ (positive) on the restricted domain
How the graph of $\sin^{-1}$ confirms an always-positive, increasing derivative

Can do

Skills

Derive $\dfrac{d}{dx}(\sin^{-1}x)$ from scratch using implicit differentiation
Apply the chain rule to differentiate $\sin^{-1}(ax)$ and $\sin^{-1}(g(x))$
State the domain of the derivative and explain why it is restricted

Key terms

$\sin^{-1}x$ (arcsine)The inverse of $\sin$ on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. Gives the angle $y$ such that $\sin y = x$, with $y \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$.

Restricted domain$\sin$ is not one-to-one on all of $\mathbb{R}$, so we restrict to $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ where it is strictly increasing and invertible.

Pythagorean identity$\sin^2 y + \cos^2 y = 1$. Rearranges to $\cos y = \pm\sqrt{1-\sin^2 y}$; we choose $+$ because $\cos y \geq 0$ on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$.

Vertical tangentOccurs where the derivative is undefined (denominator zero). The curve $y = \sin^{-1}x$ has vertical tangents at $x = \pm 1$.

Chain rule$\dfrac{d}{dx}[\sin^{-1}(u)] = \dfrac{1}{\sqrt{1-u^2}} \cdot \dfrac{du}{dx}$, where $u$ is a differentiable function of $x$.

Antiderivative connectionBecause $\dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1-x^2}}$, it follows that $\displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\,dx = \sin^{-1}x + C$.

Full derivation of $\dfrac{d}{dx}(\sin^{-1}x)$

core concept

We use the implicit differentiation method from Lesson 9.

Step 1 — Rewrite as a sine equation:

Let $y = \sin^{-1}x$. Then $x = \sin y$, restricted to $y \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$.

Step 2 — Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(x) = \frac{d}{dx}(\sin y) \qquad \Longrightarrow \qquad 1 = \cos y \cdot \frac{dy}{dx}$$

Step 3 — Solve for $\dfrac{dy}{dx}$:

$\dfrac{dy}{dx} = \dfrac{1}{\cos y}$

Step 4 — Express in terms of $x$:

Since $x = \sin y$: use $\cos^2 y = 1 - \sin^2 y = 1 - x^2$, giving $\cos y = \sqrt{1-x^2}$ (positive on the restricted domain).

$$\boxed{\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}}, \qquad x \in (-1,1)$$

Prediction check: As $x \to 1^-$, $\sqrt{1-x^2} \to 0^+$, so the derivative $\to +\infty$ — confirming the vertical tangent you predicted in the hook!

Antiderivative connection. Reversing the differentiation: $\displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\,dx = \sin^{-1}x + C$. This integral appears throughout Lesson 13 and beyond — remember this link.

$\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$. Chain rule: $\frac{d}{dx}\sin^{-1}(g(x))=\frac{g'(x)}{\sqrt{1-[g(x)]^2}}$.

Pause — copy the full derivation of $\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$ via implicit differentiation of $\sin y=x$ into your book.

Quick check: What is $\dfrac{d}{dx}(\sin^{-1}x)$?

Chain rule extension for $\sin^{-1}(g(x))$

core concept

We just saw the derivation: $y=\sin^{-1}x\Rightarrow\sin y=x\Rightarrow\cos y\frac{dy}{dx}=1\Rightarrow\frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-x^2}}$ (using $\cos y>0$ on the principal branch). That raises a question: when the argument is a composite $u=g(x)$ rather than $x$ alone, how does the chain rule modify this to give $\frac{d}{dx}\sin^{-1}(g(x))$? This card answers it → $\frac{d}{dx}\sin^{-1}(g(x))=\frac{g'(x)}{\sqrt{1-[g(x)]^2}}$.

When the argument is a function $u = g(x)$ rather than just $x$, the chain rule gives:

$$\frac{d}{dx}\!\left[\sin^{-1}(g(x))\right] = \frac{g'(x)}{\sqrt{1-[g(x)]^2}}$$

Common special case: For $\sin^{-1}(ax)$ where $a$ is a constant:

$\dfrac{d}{dx}[\sin^{-1}(ax)] = \dfrac{a}{\sqrt{1-(ax)^2}} = \dfrac{a}{\sqrt{1-a^2x^2}}$

The domain restriction becomes $-1 < ax < 1$, i.e. $-\dfrac{1}{a} < x < \dfrac{1}{a}$ (for $a > 0$).

Integration link. Reversing this: $\displaystyle\int \dfrac{a}{\sqrt{1-a^2x^2}}\,dx = \sin^{-1}(ax) + C$. More usefully: $\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\!\left(\dfrac{x}{a}\right) + C$, which is the standard HSC form you need for Lesson 13.

When the argument is a function $u = g(x)$ rather than just $x$, the chain rule gives:

Pause — copy the chain-rule extension: $\frac{d}{dx}\sin^{-1}(g(x))=\frac{g'(x)}{\sqrt{1-[g(x)]^2}}$ with a worked example into your book.

Did you get this? True or false: $\dfrac{d}{dx}[\sin^{-1}(3x)] = \dfrac{3}{\sqrt{1-9x^2}}$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BASIC DERIVATIVE

Differentiate $y = \sin^{-1}(2x)$.

Let $u = 2x$, so $y = \sin^{-1}(u)$. By the chain rule: $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-u^2}} \cdot \dfrac{du}{dx}$.

Identify the outer function ($\sin^{-1}$) and inner function ($u = 2x$).

PROBLEM 2 · COMPOSITE ARGUMENT

Find $\dfrac{d}{dx}\!\left[\sin^{-1}(x^2)\right]$.

Let $u = x^2$, so $\dfrac{du}{dx} = 2x$. Apply the chain rule: $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-u^2}} \cdot 2x$.

Identify inner function $u = x^2$; the outer is $\sin^{-1}$.

PROBLEM 3 · VERIFY ANTIDERIVATIVE

Verify that $\displaystyle\int_0^{1/2} \dfrac{1}{\sqrt{1-x^2}}\,dx = \dfrac{\pi}{6}$ by differentiating $F(x) = \sin^{-1}x$ and evaluating.

$F'(x) = \dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1-x^2}}$ ✓. So $F$ is an antiderivative of the integrand.

Confirm $F$ is the correct antiderivative.

Fill the gap: $\dfrac{d}{dx}[\sin^{-1}(5x)] = \dfrac{5}{\sqrt{1 -$ $}}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the chain rule multiplier

$\dfrac{d}{dx}[\sin^{-1}(2x)] = \dfrac{\mathbf{2}}{\sqrt{1-4x^2}}$, not $\dfrac{1}{\sqrt{1-4x^2}}$. Whenever the argument is not just $x$, you must multiply by the derivative of the inner function. Students routinely drop this factor.

Trap 02

Wrong squaring inside the root

For $\sin^{-1}(2x)$: the denominator is $\sqrt{1-(2x)^2} = \sqrt{1-4x^2}$, NOT $\sqrt{1-2x^2}$. Always square the entire inner function: $(2x)^2 = 4x^2$, $(3x)^2 = 9x^2$.

Trap 03

Wrong sign (confusing with $\cos^{-1}$)

$\dfrac{d}{dx}(\sin^{-1}x) = +\dfrac{1}{\sqrt{1-x^2}}$ (positive). $\dfrac{d}{dx}(\cos^{-1}x) = -\dfrac{1}{\sqrt{1-x^2}}$ (negative). They have the same form but opposite signs. Remember: $\sin^{-1}$ increases everywhere on its domain; $\cos^{-1}$ decreases.

Did you get this? True or false: $\dfrac{d}{dx}[\sin^{-1}(4x)] = \dfrac{4}{\sqrt{1-16x^2}}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Differentiate $y = \sin^{-1}(5x)$ and state the domain.

Find $\dfrac{d}{dx}[\sin^{-1}(\sqrt{x})]$.

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

Show that $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ by differentiating both sides and using an initial condition.

Without a calculator, find the exact gradient of $y = \sin^{-1}x$ at $x = \dfrac{\sqrt{3}}{2}$.

Odd one out: Three of these derivatives are correct. Which one is WRONG?

Revisit your thinking

Earlier you predicted what $\dfrac{d}{dx}(\sin^{-1}x)$ would look like, and whether the denominator would involve a square root.

The result is $\dfrac{1}{\sqrt{1-x^2}}$ — and as $x \to 1^-$, the denominator $\to 0$ and the derivative $\to +\infty$, exactly as the vertical tangent on the curve suggests. The Pythagorean identity $\cos^2 y + \sin^2 y = 1$ is the algebraic engine behind this result.

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Multiple choice

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Short answer

ApplyBand 31 mark

Q1. State $\dfrac{d}{dx}(\sin^{-1}x)$ and the domain on which it is valid. (1 mark)

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ApplyBand 42 marks

Q2. Find $\dfrac{d}{dx}[\sin^{-1}(3x^2)]$. (2 marks)

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AnalyseBand 53 marks

Q3. Derive $\dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1-x^2}}$ from first principles using implicit differentiation. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\dfrac{d}{dx}[\sin^{-1}(5x)] = \dfrac{5}{\sqrt{1-25x^2}}$; domain $|x| < \tfrac{1}{5}$.

2. $u = \sqrt{x} = x^{1/2}$, $u' = \dfrac{1}{2\sqrt{x}}$. So $\dfrac{d}{dx}[\sin^{-1}(\sqrt{x})] = \dfrac{1}{\sqrt{1-x}} \cdot \dfrac{1}{2\sqrt{x}} = \dfrac{1}{2\sqrt{x(1-x)}}$.

3. $\displaystyle\int_0^{1/2} \dfrac{2}{\sqrt{1-x^2}}\,dx = 2[\sin^{-1}x]_0^{1/2} = 2(\tfrac{\pi}{6} - 0) = \dfrac{\pi}{3}$.

4. $\dfrac{d}{dx}[\sin^{-1}x + \cos^{-1}x] = \dfrac{1}{\sqrt{1-x^2}} - \dfrac{1}{\sqrt{1-x^2}} = 0$, so the sum is constant. At $x=0$: $0 + \tfrac{\pi}{2} = \tfrac{\pi}{2}$. Hence $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$.

5. $x = \dfrac{\sqrt{3}}{2}$: $1-x^2 = 1 - \tfrac{3}{4} = \tfrac{1}{4}$, so $\sqrt{1-x^2} = \tfrac{1}{2}$. Gradient $= \dfrac{1}{1/2} = 2$.

Q1 (1 mark): $\dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1-x^2}}$, domain $x \in (-1,1)$ [1].

Q2 (2 marks): $u = 3x^2$, $u' = 6x$ [1]. $\dfrac{dy}{dx} = \dfrac{6x}{\sqrt{1-(3x^2)^2}} = \dfrac{6x}{\sqrt{1-9x^4}}$ [1].

Q3 (3 marks): Write $x = \sin y$ [1]. Differentiate: $1 = \cos y \cdot \dfrac{dy}{dx}$, so $\dfrac{dy}{dx} = \dfrac{1}{\cos y}$ [1]. Use $\cos y = \sqrt{1-\sin^2 y} = \sqrt{1-x^2}$ (positive on restricted domain), giving $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-x^2}}$ [1].

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← Lesson 9 · Derivatives of Inverse Functions Lesson 11 · Derivatives of cos⁻¹ x and tan⁻¹ x →

Module overview · Extension 1 · Checkpoint 2