What is first function?

Here the inner function is u = 3x, so u' = 3. Applying (d)/(dx)(u) = (u')/(1 + u^2):

What is second function?

Here u = x^2, so u' = 2x. Applying (d)/(dx)(u) = (u')/(sqrt(1 - u^2)):

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]

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 4.76 min answerUpdated 2026-05-29

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 $\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + c$ .

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What this dot point is asking
The standard derivatives
Scaled forms
Composite expressions
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 $\frac{2x}{1 + x^4}$ .

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 $y = \arcsin(x^2)$ .

First function. Here the inner function is $u = 3x$ , so $u' = 3$ . Applying $\frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}$ :

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

Second function. Here $u = x^2$ , so $u' = 2x$ . Applying $\frac{d}{dx}\arcsin(u) = \frac{u'}{\sqrt{1 - u^2}}$ :

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

Check the first 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. $\arcsin\frac{x}{5} + c$ .