What is mismatched a?

In the arctan antiderivative the factor is (1)/(a), and the argument is (x)/(a). Keep them consistent.

Differentiate y = (x^3). [2 marks]

Evaluate int(dx)/(sqrt(25 - x^2)). [2 marks]

VCAA VCAA 2022 Exam 1-style practice: Differentiate (a) y = (2x) and (b) y = \!(x4), simplifying each result.

**(a)** With u = 2x, u' = 2, and ddx(u) = u'1 + u^2: dydx = 21 + (2x)^2 = 21 + 4x^2. **(b)** Using the scaled form ddxxa = 1sqrt(a^2 - x^2) with a = 4: dydx = 1sqrt(16 - x^2). Markers reward the chain-rule factor in (a) and the correct scaled form in (b).

VCAA VCAA 2023 Exam 2-style practice: Let f(x) = x x. (a) Find f'(x). (b) Evaluate f'(1) exactly. (c) Hence find the equation of the tangent to y = f(x) at x = 1.

**(a)** Product rule with u = x, v = x: f'(x) = 1· x + x·11 + x^2 = x + x1 + x^2. **(b)** f'(1) = 1 + 11 + 1 = π4 + 12. **(c)** f(1) = 1· 1 = π4. Tangent through (1, (π)/(4)) with gradient (π)/(4) + (1)/(2): y - π4 = (π4 + 12)(x - 1). Markers reward the product rule, the exact value of f'(1), and the tangent equation.

§-Syllabus dot point

VICSpecialist MathematicsSyllabus dot point

What are the derivatives of the inverse circular functions, and how do we use the chain rule to differentiate composite expressions involving arcsin, arccos and arctan?

Differentiation of the inverse circular functions $\arcsin$ , $\arccos$ and $\arctan$ , the standard derivative results, the use of the chain rule for composite forms, and the related standard antiderivatives

A focused answer to the VCE Specialist Mathematics Unit 4 key-knowledge point on differentiating inverse circular functions. Standard derivatives of arcsin, arccos and arctan, chain-rule composites, and the corresponding antiderivatives, with a verified worked example.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The standard derivatives are $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ , $\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}$ , and $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$ . For a general scale $a > 0$ , $\frac{d}{dx}\arcsin\frac{x}{a} = \frac{1}{\sqrt{a^2 - x^2}}$ and $\frac{d}{dx}\arctan\frac{x}{a} = \frac{a}{a^2 + x^2}$ . The chain rule extends these to composites: $\frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}$ . Reversing the derivatives gives the standard antiderivatives $\int\frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + c$ and $\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + c$ .

Jump to a section

What this dot point is asking
The standard derivatives
Scaled forms
Composite expressions
Combining with the product and quotient rules
Matching antiderivatives
Examples in context
Try this

What this dot point is asking

VCAA wants you to know and use the derivatives of $\arcsin$ , $\arccos$ and $\arctan$ , to differentiate composite expressions with the chain rule, and to recognise the matching standard antiderivatives. These derivatives are the source of the inverse-trigonometric forms used in integration.

The standard derivatives

The three core results, valid where the inverse functions are differentiable, are

\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1 + x^2}.

The first two differ only in sign, consistent with the identity $\arcsin x + \arccos x = \frac{\pi}{2}$ , whose derivative is $0$ . Each can be derived by implicit differentiation: from $y = \arcsin x$ we have $\sin y = x$ , so $\cos y \frac{dy}{dx} = 1$ and $\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - x^2}}$ , taking the positive root since $\cos y \ge 0$ on the range of $\arcsin$ .

Scaled forms

Introducing a constant $a > 0$ gives the forms most useful for integration:

\frac{d}{dx}\arcsin\frac{x}{a} = \frac{1}{\sqrt{a^2 - x^2}}, \qquad \frac{d}{dx}\arctan\frac{x}{a} = \frac{a}{a^2 + x^2}.

These follow from the chain rule. For example, $\frac{d}{dx}\arcsin\frac{x}{a} = \frac{1}{\sqrt{1 - (x/a)^2}}\cdot\frac{1}{a} = \frac{1}{a\sqrt{1 - x^2/a^2}} = \frac{1}{\sqrt{a^2 - x^2}}$ .

Composite expressions

With an inner function $u = u(x)$ , the chain rule gives

\frac{d}{dx}\arcsin(u) = \frac{u'}{\sqrt{1 - u^2}}, \qquad \frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}.

So differentiating $\arctan(x^2)$ gives $\frac{2x}{1 + x^4}$ .

Combining with the product and quotient rules

In exam questions the inverse function rarely appears alone; it is usually multiplied by, or divided by, another function. A product such as $x\arctan x$ uses the product rule, differentiating $\arctan x$ to $\dfrac{1}{1 + x^2}$ while leaving the other factor, then summing. A quotient such as $\dfrac{\arcsin x}{x}$ uses the quotient rule. When the inner argument is itself a function, apply the chain rule first to get the inverse-function factor, then combine. Setting such a derivative to zero to find stationary points is a common follow-up, so keep the result in a tidy factored or common-denominator form.

Matching antiderivatives

Reversing the derivatives gives standard integrals:

\int\frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + c, \qquad \int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + c.

These are the inverse-trigonometric standard forms that appear in the integration-techniques work.

Worked example

A chain-rule composite

Differentiate $y = \arctan(3x)$ and then $y = \arcsin(x^2)$ .

Step 1: Identify the inner function for $y = \arctan(3x)$

The inner function is $u = 3x$ , so $u' = 3$ . Recognising the composite structure is essential before applying any derivative rule; the chain rule requires both the outer derivative and the inner derivative as a multiplying factor.

Step 2: Apply the chain rule to differentiate $y = \arctan(3x)$

Using $\frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}$ :

\frac{dy}{dx} = \frac{3}{1 + (3x)^2} = \frac{3}{1 + 9x^2}.

The inner derivative $u' = 3$ appears in the numerator. A common error is to write $\frac{1}{1 + 9x^2}$ and omit this factor entirely.

Step 3: Identify the inner function for $y = \arcsin(x^2)$

Here $u = x^2$ , so $u' = 2x$ . The inner function is now a power of $x$ rather than a simple linear term, so the chain-rule factor $u'$ will itself depend on $x$ .

Step 4: Apply the chain rule to differentiate $y = \arcsin(x^2)$

Using $\frac{d}{dx}\arcsin(u) = \frac{u'}{\sqrt{1 - u^2}}$ :

\frac{dy}{dx} = \frac{2x}{\sqrt{1 - (x^2)^2}} = \frac{2x}{\sqrt{1 - x^4}}.

Check the first result at $x = 0$ : the derivative is $\frac{3}{1} = 3$ , which matches the gradient of $\arctan(3x)$ near the origin where $\arctan(3x) \approx 3x$ . Consistent.

Examples in context

Example 1. $\frac{d}{dx}\arcsin\frac{x}{4} = \frac{1}{\sqrt{16 - x^2}}$ .

Example 2. $\int\frac{dx}{9 + x^2} = \frac{1}{3}\arctan\frac{x}{3} + c$ , with $a = 3$ .

Try this

Q1. Differentiate $y = \arcsin x$ . [1 mark]

Cue. $\frac{1}{\sqrt{1 - x^2}}$ .

Q2. Differentiate $y = \arctan(x^3)$ . [2 marks]

Cue. $\frac{3x^2}{1 + x^6}$ .

Q3. Evaluate $\int\frac{dx}{\sqrt{25 - x^2}}$ . [2 marks]

Cue. $\arcsin\frac{x}{5} + c$ .

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

VCAA 2022 Exam 13 marksDifferentiate (a)

y = \arctan(2x)

and (b)

y = \arcsin\!\left(\dfrac{x}{4}\right)

, simplifying each result.

Show worked answer →

(a) With $u = 2x$ , $u' = 2$ , and $\dfrac{d}{dx}\arctan(u) = \dfrac{u'}{1 + u^2}$ :
$\dfrac{dy}{dx} = \dfrac{2}{1 + (2x)^2} = \dfrac{2}{1 + 4x^2}$ .

(b) Using the scaled form $\dfrac{d}{dx}\arcsin\dfrac{x}{a} = \dfrac{1}{\sqrt{a^2 - x^2}}$ with $a = 4$ :
$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{16 - x^2}}$ .

Markers reward the chain-rule factor in (a) and the correct scaled form in (b).

VCAA 2023 Exam 25 marksLet

f(x) = x\arctan x

. (a) Find

f'(x)

. (b) Evaluate

f'(1)

exactly. (c) Hence find the equation of the tangent to

y = f(x)

x = 1

Show worked answer →

(a) Product rule with $u = x$ , $v = \arctan x$ : $f'(x) = 1\cdot\arctan x + x\cdot\dfrac{1}{1 + x^2} = \arctan x + \dfrac{x}{1 + x^2}$ .

(b) $f'(1) = \arctan 1 + \dfrac{1}{1 + 1} = \dfrac{\pi}{4} + \dfrac{1}{2}$ .

(c) $f(1) = 1\cdot\arctan 1 = \dfrac{\pi}{4}$ . Tangent through $\left(1, \frac{\pi}{4}\right)$ with gradient $\frac{\pi}{4} + \frac{1}{2}$ :
$y - \dfrac{\pi}{4} = \left(\dfrac{\pi}{4} + \dfrac{1}{2}\right)(x - 1)$ .

Markers reward the product rule, the exact value of $f'(1)$ , and the tangent equation.