How do you differentiate arccos of 4x?

With u equals 4x, the derivative is minus four over the square root of one minus sixteen x squared.

What is the derivative of arctan of x over a?

a over a squared plus x squared.

How do you differentiate arccos of 4x?

With u equals 4x, the derivative is minus four over the square root of one minus sixteen x squared.

What is the derivative of arctan of x over a?

a over a squared plus x squared.

Year 12 Maths Extension 1 (2027) Calculus

Derivatives of inverse trig functions

Q: What is the derivative of arcsin, arccos and arctan?

One over the square root of one minus x squared; minus that; and one over one plus x squared.

Q: Why does the arccos derivative have a minus sign?

Because arcsin x plus arccos x is the constant pi over 2, so their derivatives are equal and opposite.

Q: Do these need the chain rule?

Yes when there is an inner function; multiply by u prime and replace x by u.

20 practice questions 2 video lessons Theory + worked examples

Learn the derivatives of inverse trigonometric functions for NSW Year 12 Mathematics Extension 1. Arcsin, arccos and arctan each have a standard derivative, proved from the inverse-function rule.

You will learn to differentiate inverse trig functions of more complex expressions using the chain rule, and to combine them with the product and quotient rules to find tangents and gradients — an essential calculus skill in Extension 1.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

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Watch 2 video(s)

Proofs of derivatives of Inverse Trigonometric Functions Watch
Calculating derivatives of inverse trigonometric Functions Watch

Theory

The inverse trig functions have standard derivatives: $\frac{d}{d x} \sin^{- 1} x = \frac{1}{\sqrt{1 - x^{2}}}$ , $\frac{d}{d x} \cos^{- 1} x = - \frac{1}{\sqrt{1 - x^{2}}}$ , $\frac{d}{d x} \tan^{- 1} x = \frac{1}{1 + x^{2}}$ . This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-04) applies them with the chain rule.

The inverse trigonometric functions have standard derivatives worth memorising — they are the reverse of the standard integrals that give $\sin^{- 1}$ and $\tan^{- 1}$ .

$\frac{d}{d x} \sin^{- 1} x = \frac{1}{\sqrt{1 - x^{2}}}$ , $\frac{d}{d x} \cos^{- 1} x = - \frac{1}{\sqrt{1 - x^{2}}}$ (the $\cos^{- 1}$ form differs only by a minus sign), and $\frac{d}{d x} \tan^{- 1} x = \frac{1}{1 + x^{2}}$ .

With an inner function $u = f (x)$ , apply the chain rule — multiply by $u^{'}$ and replace $x$ by $u$ inside the root.

NESA link. Part of the Year 12 Calculus topic, outcome ME1-12-04 ("selects and applies differentiation and integration techniques to solve problems") with MAO-WM-01.

y = \sin^{- 1} x

: its gradient

\frac{1}{\sqrt{1 - x^{2}}}

grows without bound near

x = \pm 1

y = \tan^{- 1} x

: its gradient

\frac{1}{1 + x^{2}}

is largest at the origin.

\frac{d}{d x} \sin^{- 1} x = \frac{1}{\sqrt{1 - x^{2}}}, \frac{d}{d x} \cos^{- 1} x = - \frac{1}{\sqrt{1 - x^{2}}}, \frac{d}{d x} \tan^{- 1} x = \frac{1}{1 + x^{2}} .

d/dx arcsin x = 1/sqrt(1-x^2); d/dx arccos x = -1/sqrt(1-x^2); d/dx arctan x = 1/(1+x^2)

With an inner function (chain rule):

\frac{d}{d x} \sin^{- 1} u = \frac{u^{'}}{\sqrt{1 - u^{2}}}, \frac{d}{d x} \tan^{- 1} u = \frac{u^{'}}{1 + u^{2}} .

d/dx arcsin u = u' / sqrt(1-u^2); d/dx arctan u = u'/(1+u^2)

Useful scaled forms

Function	Derivative
$\sin^{- 1} \frac{x}{a}$	$\frac{1}{\sqrt{a^{2} - x^{2}}}$
$\tan^{- 1} \frac{x}{a}$	$\frac{a}{a^{2} + x^{2}}$

How to differentiate an inverse trig function

Identify the inner function $u$ and compute $u^{'}$ .
Apply the standard form, replacing $x$ by $u$ inside the root or the $1 + u^{2}$ , and keep the minus sign for $\cos^{- 1}$ .
Multiply by $u^{'}$ (chain rule).
For a value or tangent, substitute the $x$ -value; the tangent uses $y - y_{1} = m (x - x_{1})$ .

Example 1 — Chain rule (cosine)

Differentiate

y = \cos^{- 1} (4 x)

Solution

$u = 4 x$ , $u^{'} = 4$ ; use $- \frac{u^{'}}{\sqrt{1 - u^{2}}}$ .

\frac{d y}{d x}

=

- \frac{4}{\sqrt{1 - (4 x)^{2}}} = - \frac{4}{\sqrt{1 - 16 x^{2}}}

d/dx arccos(4x) = -4/sqrt(1-16x^2)

So $\frac{d y}{d x} = - \frac{4}{\sqrt{1 - 16 x^{2}}}$ .

Example 2 — Scaled argument (tangent)

Differentiate

y = \tan^{- 1} \frac{x}{4}

Solution

Use $\frac{d}{d x} \tan^{- 1} \frac{x}{a} = \frac{a}{a^{2} + x^{2}}$ with $a = 4$ .

\frac{d y}{d x}

=

\frac{4}{16 + x^{2}}

d/dx arctan(x/4) = 4/(16 + x^2)

So $\frac{d y}{d x} = \frac{4}{16 + x^{2}}$ .

Example 3 — A gradient value

Find the gradient of

y = \sin^{- 1} \frac{x}{2}

x = 1

Solution

$\frac{d}{d x} \sin^{- 1} \frac{x}{a} = \frac{1}{\sqrt{a^{2} - x^{2}}}$ , $a = 2$ .

$\frac{d y}{d x}$	$=$	$\frac{1}{\sqrt{4 - x^{2}}}$
$x = 1$	$\Rightarrow$	$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

gradient at x=1 is 1/sqrt(3) = sqrt(3)/3

Gradient: $\frac{\sqrt{3}}{3}$ .

Example 4 — Tangent to arctan

Find the tangent to

y = \tan^{- 1} x

x = 1

Solution

$\frac{d y}{d x} = \frac{1}{1 + x^{2}}$ , so at $x = 1$ , $m = \frac{1}{2}$ and $y = \frac{π}{4}$ .

y - \frac{π}{4}

=

\frac{1}{2} (x - 1)

tangent: y = (1/2)(x - 1) + pi/4

Tangent: $y = \frac{1}{2} (x - 1) + \frac{π}{4}$ .

Common pitfalls

The minus sign. The

\cos^{- 1}

derivative carries a minus sign:

- \frac{1}{\sqrt{1 - x^{2}}}

Chain rule. With an inner function, multiply by

u^{'}

— and the root is

\sqrt{1 - u^{2}}

, not

\sqrt{1 - x^{2}}

Scaled tangent. For

\tan^{- 1} (k x)

the derivative is

\frac{k}{1 + k^{2} x^{2}}

, not

\frac{1}{1 + k^{2} x^{2}}

Domain.

\sin^{- 1}

and

\cos^{- 1}

need \(-1

Frequently asked questions

What is the derivative of arcsin, arccos and arctan?

$\frac{1}{\sqrt{1 - x^{2}}}$ , $- \frac{1}{\sqrt{1 - x^{2}}}$ and $\frac{1}{1 + x^{2}}$ respectively.

How do you differentiate arccos(4x)?

With $u = 4 x$ , $\frac{d}{d x} \cos^{- 1} (4 x) = - \frac{4}{\sqrt{1 - 16 x^{2}}}$ .

What is the derivative of arctan(x/a)?

$\frac{a}{a^{2} + x^{2}}$ .

Why does the arccos derivative have a minus sign?

Because $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$ is constant, so their derivatives are equal and opposite.

Do these need the chain rule?

Yes when there is an inner function — multiply by $u^{'}$ and replace $x$ by $u$ .

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Integration giving inverse trig functions

Derivatives of inverse trig functions

📖 Theory

Useful scaled forms

How to differentiate an inverse trig function

Common pitfalls

Frequently asked questions

Theory