Math Core

The inverse trig functions $\arcsin x$ , $\arccos x$ and $\arctan x$ answer the question "which angle has this sine, cosine or tangent?" Their derivatives are a surprise: they contain no trig functions at all, just algebraic expressions. That fact becomes very useful later, when you run the process backward to find antiderivatives.

A quick review of the inverse trig functions

Sine, cosine and tangent aren't one-to-one, so each is restricted to an interval before it's inverted:

function	domain	range (output angles)
$y = \arcsin x$	$-1 \le x \le 1$	$-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$
$y = \arccos x$	$-1 \le x \le 1$	$0 \le y \le \pi$
$y = \arctan x$	all real $x$	$-\dfrac{\pi}{2} \lt y \lt \dfrac{\pi}{2}$

You'll also see the notation $\sin^{-1} x$ for $\arcsin x$ . As always, the $-1$ means "inverse," not reciprocal.

Deriving the derivative of arcsin x

Let $y = \arcsin x$ . Then $\sin y = x$ , with $y$ between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ . Differentiate implicitly:

\cos y \cdot \frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{\cos y}.

To write this in terms of $x$ , use $\cos^2 y + \sin^2 y = 1$ , so $\cos y = \pm\sqrt{1 - \sin^2 y} = \pm\sqrt{1 - x^2}$ . Because $y$ is between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ , $\cos y \ge 0$ , so we take the positive root:

\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}.

The same method works for arctangent. From $\tan y = x$ you get $\sec^2 y \cdot \dfrac{dy}{dx} = 1$ , and $\sec^2 y = 1 + \tan^2 y = 1 + x^2$ . So $\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^2}$ .

Derivatives of the inverse trig functions

\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}

With the chain rule, for an inner function $u = g(x)$ :

\frac{d}{dx}\arcsin u = \frac{u'}{\sqrt{1 - u^2}} \qquad \frac{d}{dx}\arccos u = -\frac{u'}{\sqrt{1 - u^2}} \qquad \frac{d}{dx}\arctan u = \frac{u'}{1 + u^2}

The derivative of $\arccos x$ is the negative of the derivative of $\arcsin x$ . That makes sense: $\arcsin x + \arccos x = \dfrac{\pi}{2}$ for every $x$ in $[-1, 1]$ , and the derivative of a constant is 0.

For completeness, the other three are $\dfrac{d}{dx}\operatorname{arccot} x = -\dfrac{1}{1 + x^2}$ , $\dfrac{d}{dx}\operatorname{arcsec} x = \dfrac{1}{|x|\sqrt{x^2 - 1}}$ and $\dfrac{d}{dx}\operatorname{arccsc} x = -\dfrac{1}{|x|\sqrt{x^2 - 1}}$ . The AP exam focuses on arcsine, arccosine and arctangent.

Reading the graph

The graph of $y = \arctan x$ always rises, and it is steepest at the origin. The formula agrees: $\dfrac{1}{1 + x^2}$ is always positive, equals 1 at $x = 0$ , and shrinks toward 0 as $|x|$ grows. At $x = 1$ the slope is $\dfrac{1}{2}$ .

y = arctan x with its tangent line at (1, π/4). The slope there is 1/(1 + 1²) = 1/2.Open in grapher →

Examples

Worked example: Chain rule with arcsin and arctan

Differentiate each function.

$y = \arcsin(3x)$
$y = \arctan(x^2)$

Solutions.

Here $u = 3x$ and $u' = 3$ : $\;y' = \dfrac{3}{\sqrt{1 - (3x)^2}} = \dfrac{3}{\sqrt{1 - 9x^2}}$ .
Here $u = x^2$ and $u' = 2x$ : $\;y' = \dfrac{2x}{1 + (x^2)^2} = \dfrac{2x}{1 + x^4}$ .

Common mistake

Square the whole inner function. In part 1, $(3x)^2 = 9x^2$ , not $3x^2$ . In part 2, $(x^2)^2 = x^4$ , not $x^2$ .

Worked example: A value of the derivative

Let $f(x) = \arccos\!\left(\dfrac{x}{2}\right)$ . Find $f'(1)$ .

Solution. With $u = \dfrac{x}{2}$ and $u' = \dfrac{1}{2}$ ,

f'(x) = -\frac{1/2}{\sqrt{1 - x^2/4}}, \qquad f'(1) = -\frac{1/2}{\sqrt{3/4}} = -\frac{1/2}{\sqrt{3}/2} = -\frac{1}{\sqrt{3}}.

Worked example: Combining rules

Differentiate $y = x\arctan x$ .

Solution. Use the product rule:

y' = 1 \cdot \arctan x + x \cdot \frac{1}{1 + x^2} = \arctan x + \frac{x}{1 + x^2}.

Tip

If you forget a formula, rebuild it in 30 seconds: write $\sin y = x$ (or $\tan y = x$ ), differentiate implicitly, and use a Pythagorean identity to get back to $x$ .

Practice

Practice 1

Find $\dfrac{dy}{dx}$ if $y = \arctan(5x)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

Find $\dfrac{dy}{dx}$ if $y = \arcsin\!\left(\dfrac{x}{4}\right)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

Find $\dfrac{dy}{dx}$ if $y = \arctan(e^x)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

Find $\dfrac{dy}{dx}$ if $y = \arccos(x^2)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 5

Let $f(x) = \arctan(2x)$ . Find $f'\!\left(\dfrac{1}{2}\right)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

(AP-style) If $y = \arcsin(\sqrt{x})$ , then $\dfrac{dy}{dx} =$

Practice 7

Write an equation of the line tangent to $y = \arctan x$ at $x = 1$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 8

(AP-style) Let $f(x) = x\arcsin x$ . What is $f'\!\left(\dfrac{1}{2}\right)$ ?