Trigonometric and Inverse Trigonometric Functions

Lessons

Senior 5 · Mathematics

Trigonometric and Inverse Trigonometric Functions

5 lessons

Lesson 4 of 5

Derivatives of inverse trigonometric functions, and successive derivatives

25 min

The slope of an inverse

The graph of $y = \arcsin x$ is the graph of $y = \sin x$ reflected in the line $y = x$ (Lesson 1). Reflection swaps rise and run, so where the sine has slope $m$ , the inverse has slope $\frac{1}{m}$ . This one idea gives the derivatives of all the inverse trigonometric functions, and they turn out to be algebraic: no trigonometric function is left in the answer. That is why they appear again and again in S6 integration.

Important

Derivative of $\arcsin$ . Let $y = \arcsin x$ with $-1 \lt x \lt 1$ . Then $\sin y = x$ and $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . Differentiate both sides with respect to $x$ (chain rule on the left):

\cos y\cdot\frac{dy}{dx} = 1 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1 - x^2}}.

The positive root is right because $\cos y \gt 0$ for $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . In the same way, with $\tan y = x$ and $\sec^2 y = 1 + \tan^2 y = 1 + x^2$ :

(\arctan x)' = \frac{1}{1 + x^2}.

$f (x)$	$f^{'} (x)$	valid for	with the chain rule
$\arcsin x$	$\frac{1}{\sqrt{1 - x^2}}$	$-1 \lt x \lt 1$	$\frac{u'}{\sqrt{1 - u^2}}$
$\arccos x$	$-\frac{1}{\sqrt{1 - x^2}}$	$-1 \lt x \lt 1$	$-\frac{u'}{\sqrt{1 - u^2}}$
$\arctan x$	$\frac{1}{1 + x^2}$	all $x$	$\frac{u'}{1 + u^2}$
$\operatorname{arccot} x$	$-\frac{1}{1 + x^2}$	all $x$	$-\frac{u'}{1 + u^2}$

The derivative of $\arccos$ is minus that of $\arcsin$ , as it must be: $\arcsin x + \arccos x = \frac{\pi}{2}$ is constant. At $x = \pm 1$ the graphs of $\arcsin$ and $\arccos$ have vertical tangents, so they are not differentiable there.

Figure 1.

\arctan

is steepest at the origin (slope 1) and flattens out towards its asymptotes, so its derivative

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

peaks at 0 and tends to 0.

Worked example 1: inverse functions with the chain rule

Differentiate (a) $\arcsin 3x$ (b) $\arctan(x^2)$ (c) $x\arccos x - \sqrt{1 - x^2}$ .

Solution

$\frac{3}{\sqrt{1 - 9x^2}}$ , for $|x| \lt \frac{1}{3}$ .
$\frac{2x}{1 + x^4}$ .
$\arccos x + x\left(-\frac{1}{\sqrt{1 - x^2}}\right) - \frac{-2x}{2\sqrt{1 - x^2}} = \arccos x - \frac{x}{\sqrt{1 - x^2}} + \frac{x}{\sqrt{1 - x^2}} = \arccos x$ .

Worked example 2: a derivative that reveals an identity

Differentiate $y = \arctan\frac{1 + x}{1 - x}$ for $x \lt 1$ , and comment.

Solution With $u = \frac{1 + x}{1 - x}$ , $u' = \frac{(1 - x) + (1 + x)}{(1 - x)^2} = \frac{2}{(1 - x)^2}$ and $1 + u^2 = \frac{(1 - x)^2 + (1 + x)^2}{(1 - x)^2} = \frac{2(1 + x^2)}{(1 - x)^2}$ . So

y' = \frac{u'}{1 + u^2} = \frac{2}{(1 - x)^2}\cdot\frac{(1 - x)^2}{2(1 + x^2)} = \frac{1}{1 + x^2},

the same as $(\arctan x)'$ . So $y - \arctan x$ is constant on $x \lt 1$ ; at $x = 0$ it is $\arctan 1 - 0 = \frac{\pi}{4}$ . Hence $\arctan\frac{1 + x}{1 - x} = \arctan x + \frac{\pi}{4}$ for $x \lt 1$ , which is the addition formula of Unit 1 in disguise.

Definition

Successive (higher-order) derivatives: the second derivative $f^{''} = (f^{'})^{'}$ , the third $f^{'''} = (f^{''})^{'}$ , and in general the $n$ th derivative $f^{(n)} = \left(f^{(n-1)}\right)'$ , also written $\frac{d^ny}{dx^n}$ .

Example: for $f(x) = \sin x$ : $f'(x) = \cos x$ , $f''(x) = -\sin x$ , $f'''(x) = -\cos x$ , $f^{(4)}(x) = \sin x$ .

Patterns in successive derivatives

Since $(\sin x)' = \cos x = \sin\left(x + \frac{\pi}{2}\right)$ , each derivative of a sine adds $\frac{\pi}{2}$ to the angle (Figure 2). So, for every whole number $n$ ,

(\sin x)^{(n)} = \sin\left(x + \frac{n\pi}{2}\right), \qquad (\cos x)^{(n)} = \cos\left(x + \frac{n\pi}{2}\right), \qquad \left(\sin(ax + b)\right)^{(n)} = a^n\sin\left(ax + b + \frac{n\pi}{2}\right).

Figure 2. Differentiating

\sin x

four times returns to

\sin x

: each step adds

\frac{\pi}{2}

to the angle, since

\cos x = \sin\left(x + \frac{\pi}{2}\right)

Worked example 3: a high derivative

For $f(x) = \sin 2x$ , find $f^{(27)}(0)$ .

Solution $f^{(27)}(x) = 2^{27}\sin\left(2x + \frac{27\pi}{2}\right)$ . At $x = 0$ : $\frac{27\pi}{2} = 13\pi + \frac{\pi}{2}$ , and $\sin\left(13\pi + \frac{\pi}{2}\right) = -\sin\frac{\pi}{2} = -1$ (an odd multiple of $π \pi$ changes the sign). So $f^{(27)}(0) = -2^{27} = -134\,217\,728$ .

Worked example 4: equations linking derivatives

(a) Show that $y = A\cos\omega t + B\sin\omega t$ satisfies $\frac{d^2y}{dt^2} + \omega^2 y = 0$ . (b) Show that $y = \arctan x$ satisfies $(1 + x^{2}) y^{''} + 2 x y^{'} = 0$ .

Solution

$y' = -A\omega\sin\omega t + B\omega\cos\omega t$ , $y'' = -A\omega^2\cos\omega t - B\omega^2\sin\omega t = -\omega^2 y$ . This is the equation of simple harmonic motion (Lesson 5).
$(1 + x^{2}) y^{'} = 1$ . Differentiate both sides with the product rule: $2 x y^{'} + (1 + x^{2}) y^{''} = 0$ .

Warning

Mind the domain. $(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}$ only for $-1 \lt x \lt 1$ ; at $x = \pm 1$ the derivative does not exist. In the chain rule for $\arcsin u$ , square the whole inner function: $(\arcsin 3x)' = \frac{3}{\sqrt{1 - 9x^2}}$ , not $\frac{3}{\sqrt{1 - 3x^2}}$ .

Tip

Remember one, derive the rest. Learn $(\arcsin x)'$ and $(\arctan x)'$ . Each "arc-co" function has the same derivative with a minus sign, because $\arccos x = \frac{\pi}{2} - \arcsin x$ and $\operatorname{arccot} x = \frac{\pi}{2} - \arctan x$ .

Khan Academy: Derivative of inverse sine. Watch how writing y = arcsin x as sin y = x and differentiating both sides gives 1 over the square root of 1 − x².

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Practice

The last two are harder.

Differentiate (a) $\arcsin\frac{x}{2}$ (b) $\arctan 5x$ (c) $\arccos 2x$ (d) $x^2\arctan x$ .
Differentiate $\arcsin\sqrt{x}$ for $0 \lt x \lt 1$ .
Show that $\frac{d}{dx}(\arcsin x + \arccos x) = 0$ , and explain what this means.
Prove that $(\arctan x)' = \frac{1}{1 + x^2}$ .
Find $f^{'''} (x)$ for $f(x) = x\sin x$ .
Write the $n$ th derivative of $\cos 3x$ , and find the 10th derivative at $x = 0$ .
Show that $y = \sin 2x + 3\cos 2x$ satisfies $y^{''} + 4 y = 0$ .
Show that $y = \arcsin x$ satisfies $(1 - x^{2}) y^{''} - x y^{'} = 0$ for $|x| \lt 1$ .

Show answersHide answers

(a) $\frac{\frac{1}{2}}{\sqrt{1 - \frac{x^2}{4}}} = \frac{1}{\sqrt{4 - x^2}}$ (b) $\frac{5}{1 + 25x^2}$ (c) $-\frac{2}{\sqrt{1 - 4x^2}}$ (d) $2x\arctan x + \frac{x^2}{1 + x^2}$ .
$\frac{\frac{1}{2\sqrt{x}}}{\sqrt{1 - x}} = \frac{1}{2\sqrt{x - x^2}}$ .
$\frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - x^2}} = 0$ : the sum is constant on $(- 1, 1)$ ; its value is $\frac{\pi}{2}$ (put $x = 0$ ).
$y = \arctan x \Rightarrow \tan y = x \Rightarrow \sec^2 y\cdot y' = 1 \Rightarrow y' = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$ .
$f' = \sin x + x\cos x$ , $f'' = 2\cos x - x\sin x$ , $f''' = -3\sin x - x\cos x$ .
$3^n\cos\left(3x + \frac{n\pi}{2}\right)$ ; at $x = 0$ , $n = 10$ : $3^{10}\cos 5\pi = -59\,049$ .
$y'' = -4\sin 2x - 12\cos 2x = -4y$ .
$y' = (1 - x^2)^{-\frac{1}{2}}$ , $y'' = x(1 - x^2)^{-\frac{3}{2}}$ , so $(1 - x^2)y'' = x(1 - x^2)^{-\frac{1}{2}} = xy'$ .

Sources and credits

Written by Ganzaa: the explanations, the derivations, the diagrams, the worked examples, the practice items and the quiz questions.
Video: Derivative of inverse sine, Khan Academy, embedded from YouTube.

1The slope of an inverse

2Patterns in successive derivatives

The slope of an inverse

Patterns in successive derivatives