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Differentiation of inverse circular functions: VCE Specialist Mathematics Unit 4

Syllabus dot point

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

7 min read

Quick answer

The standard derivatives are ddxarcsin⁡x=11−x2\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, ddxarccos⁡x=−11−x2\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, and ddxarctan⁡x=11+x2\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}. For a general scale a>0a > 0, ddxarcsin⁡xa=1a2−x2\frac{d}{dx}\arcsin\frac{x}{a} = \frac{1}{\sqrt{a^2 - x^2}} and ddxarctan⁡xa=aa2+x2\frac{d}{dx}\arctan\frac{x}{a} = \frac{a}{a^2 + x^2}. The chain rule extends these to composites: ddxarctan⁡(u)=u′1+u2\frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}. Reversing the derivatives gives the standard antiderivatives ∫dxa2−x2=arcsin⁡xa+c\int\frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + c and ∫dxa2+x2=1aarctan⁡xa+c\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + c.

Jump to a section
  1. What this dot point is asking
  2. The standard derivatives
  3. Scaled forms
  4. Composite expressions
  5. Combining with the product and quotient rules
  6. Matching antiderivatives
  7. Examples in context
  8. Try this
  9. Exam-style questions

What this dot point is asking

VCAA wants you to know and use the derivatives of arcsin⁡\arcsin, arccos⁡\arccos and arctan⁡\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

ddxarcsin⁡x=11−x2,ddxarccos⁡x=−11−x2,ddxarctan⁡x=11+x2.\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=π2\arcsin x + \arccos x = \frac{\pi}{2}, whose derivative is 00. Each can be derived by implicit differentiation: from y=arcsin⁡xy = \arcsin x we have sin⁡y=x\sin y = x, so cos⁡ydydx=1\cos y \frac{dy}{dx} = 1 and dydx=1cos⁡y=11−x2\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - x^2}}, taking the positive root since cos⁡y≥0\cos y \ge 0 on the range of arcsin⁡\arcsin.

Scaled forms

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

ddxarcsin⁡xa=1a2−x2,ddxarctan⁡xa=aa2+x2.\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, ddxarcsin⁡xa=11−(x/a)2⋅1a=1a1−x2/a2=1a2−x2\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)u = u(x), the chain rule gives

ddxarcsin⁡(u)=u′1−u2,ddxarctan⁡(u)=u′1+u2.\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⁡(x2)\arctan(x^2) gives 2x1+x4\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 xarctan⁡xx\arctan x uses the product rule, differentiating arctan⁡x\arctan x to 11+x2\dfrac{1}{1 + x^2} while leaving the other factor, then summing. A quotient such as arcsin⁡xx\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:

∫dxa2−x2=arcsin⁡xa+c,∫dxa2+x2=1aarctan⁡xa+c.\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)y = \arctan(3x) and then y=arcsin⁡(x2)y = \arcsin(x^2).

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

The inner function is u=3xu = 3x, so u′=3u' = 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)y = \arctan(3x)

Using ddxarctan⁡(u)=u′1+u2\frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}:

dydx=31+(3x)2=31+9x2.\frac{dy}{dx} = \frac{3}{1 + (3x)^2} = \frac{3}{1 + 9x^2}.

The inner derivative u′=3u' = 3 appears in the numerator. A common error is to write 11+9x2\frac{1}{1 + 9x^2} and omit this factor entirely.

Step 3: Identify the inner function for y=arcsin⁡(x2)y = \arcsin(x^2)

Here u=x2u = x^2, so u′=2xu' = 2x. The inner function is now a power of xx rather than a simple linear term, so the chain-rule factor u′u' will itself depend on xx.

Step 4: Apply the chain rule to differentiate y=arcsin⁡(x2)y = \arcsin(x^2)

Using ddxarcsin⁡(u)=u′1−u2\frac{d}{dx}\arcsin(u) = \frac{u'}{\sqrt{1 - u^2}}:

dydx=2x1−(x2)2=2x1−x4.\frac{dy}{dx} = \frac{2x}{\sqrt{1 - (x^2)^2}} = \frac{2x}{\sqrt{1 - x^4}}.

Check the first result at x=0x = 0: the derivative is 31=3\frac{3}{1} = 3, which matches the gradient of arctan⁡(3x)\arctan(3x) near the origin where arctan⁡(3x)≈3x\arctan(3x) \approx 3x. Consistent.

Common exam traps
Sign of the arccos derivative
ddxarccos⁡x=−11−x2\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, with a minus sign. Dropping it is a frequent slip.
Forgetting the chain rule factor
ddxarctan⁡(3x)\frac{d}{dx}\arctan(3x) is 31+9x2\frac{3}{1 + 9x^2}, not 11+9x2\frac{1}{1 + 9x^2}. The inner derivative multiplies.
Confusing the two standard antiderivatives
1a2−x2\frac{1}{\sqrt{a^2 - x^2}} integrates to arcsin⁡\arcsin; 1a2+x2\frac{1}{a^2 + x^2} integrates to 1aarctan⁡\frac{1}{a}\arctan. The square root distinguishes them.
Mismatched aa
In the arctan antiderivative the factor is 1a\frac{1}{a}, and the argument is xa\frac{x}{a}. Keep them consistent.

Examples in context

Example 1. ddxarcsin⁡x4=116−x2\frac{d}{dx}\arcsin\frac{x}{4} = \frac{1}{\sqrt{16 - x^2}}.

Example 2. ∫dx9+x2=13arctan⁡x3+c\int\frac{dx}{9 + x^2} = \frac{1}{3}\arctan\frac{x}{3} + c, with a=3a = 3.

Try this

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

  • Cue. 11−x2\frac{1}{\sqrt{1 - x^2}}.

Q2. Differentiate y=arctan⁡(x3)y = \arctan(x^3). [2 marks]

  • Cue. 3x21+x6\frac{3x^2}{1 + x^6}.

Q3. Evaluate ∫dx25−x2\int\frac{dx}{\sqrt{25 - x^2}}. [2 marks]

  • Cue. arcsin⁡x5+c\arcsin\frac{x}{5} + c.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks
Differentiate (a) y=arctan⁡(2x)y = \arctan(2x) and (b) y=arcsin⁡ ⁣(x4)y = \arcsin\!\left(\dfrac{x}{4}\right), simplifying each result.
Show worked answer →

(a) With u=2xu = 2x, u′=2u' = 2, and ddxarctan⁡(u)=u′1+u2\dfrac{d}{dx}\arctan(u) = \dfrac{u'}{1 + u^2}:
dydx=21+(2x)2=21+4x2\dfrac{dy}{dx} = \dfrac{2}{1 + (2x)^2} = \dfrac{2}{1 + 4x^2}.

(b) Using the scaled form ddxarcsin⁡xa=1a2−x2\dfrac{d}{dx}\arcsin\dfrac{x}{a} = \dfrac{1}{\sqrt{a^2 - x^2}} with a=4a = 4:
dydx=116−x2\dfrac{dy}{dx} = \dfrac{1}{\sqrt{16 - x^2}}.

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

Practice5 marks
Let f(x)=xarctan⁡xf(x) = x\arctan x. (a) Find f′(x)f'(x). (b) Evaluate f′(1)f'(1) exactly. (c) Hence find the equation of the tangent to y=f(x)y = f(x) at x=1x = 1.
Show worked answer →

(a) Product rule with u=xu = x, v=arctan⁡xv = \arctan x: f′(x)=1⋅arctan⁡x+x⋅11+x2=arctan⁡x+x1+x2f'(x) = 1\cdot\arctan x + x\cdot\dfrac{1}{1 + x^2} = \arctan x + \dfrac{x}{1 + x^2}.

(b) f′(1)=arctan⁡1+11+1=π4+12f'(1) = \arctan 1 + \dfrac{1}{1 + 1} = \dfrac{\pi}{4} + \dfrac{1}{2}.

(c) f(1)=1⋅arctan⁡1=π4f(1) = 1\cdot\arctan 1 = \dfrac{\pi}{4}. Tangent through (1,π4)\left(1, \frac{\pi}{4}\right) with gradient π4+12\frac{\pi}{4} + \frac{1}{2}:
y−π4=(π4+12)(x−1)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)f'(1), and the tangent equation.

Practise this

Sources & how we know this

Based on the official VCAA study design and published VCAA past papers and marking guidelines. Not yet individually reviewed by a person; always check VCAA and your school's materials for current rules.

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