Derivatives of Inverse Trig Functions

The inverse trig functions answer questions like “which angle has a sine of $0.5$?” Their derivatives are surprising: there’s no trig left in them at all, just algebraic expressions like $\dfrac{1}{1 + x^2}$. That makes them important later, when you need to undo derivatives. All angles here are in radians.

Key ideas

Notation: arcsin x or sin⁻¹ x

The inverse sine is written two ways, and both mean the same thing:

$$\arcsin x = \sin^{-1} x$$

Either one means “the angle (in radians) whose sine is $x$”. The $-1$ means inverse, not reciprocal: $\sin^{-1} x$ is not $\dfrac{1}{\sin x}$ (that’s $\csc x$). The same goes for $\arccos x = \cos^{-1} x$ and $\arctan x = \tan^{-1} x$. This page mostly uses the “arc” names to avoid confusion, but you’ll see both on the AP exam.

Restricted ranges

Sine, cosine, and tangent aren’t one-to-one, so each is restricted to an interval where it is, and the inverse takes its outputs from that interval:

Function	Domain	Range (output angles)
$\arcsin x$	$-1 \le x \le 1$	$-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$
$\arccos x$	$-1 \le x \le 1$	$0 \le y \le \pi$
$\arctan x$	all real $x$	$-\dfrac{\pi}{2} \lt y \lt \dfrac{\pi}{2}$

For example, $\arcsin\dfrac{1}{2} = \dfrac{\pi}{6}$, $\arccos 0 = \dfrac{\pi}{2}$, and $\arctan 1 = \dfrac{\pi}{4}$. (If you learned special angles in degrees, remember $\dfrac{\pi}{6} = 30^\circ$, $\dfrac{\pi}{4} = 45^\circ$, $\dfrac{\pi}{2} = 90^\circ$.)

Where the arcsin derivative comes from

Let $y = \arcsin x$. Then $\sin y = x$. Use implicit differentiation:

$$\cos y \frac{dy}{dx} = 1 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{1}{\cos y}$$

Now write $\cos y$ in terms of $x$. Since $\sin^2 y + \cos^2 y = 1$, $\cos y = \pm\sqrt{1 - \sin^2 y} = \pm\sqrt{1 - x^2}$. On the range $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$, cosine is never negative, so take the positive root:

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

This is the same idea as the derivative of an inverse function: the slope of the inverse is the reciprocal of the slope of sine at the matching point.

Where the arctan derivative comes from

Let $y = \arctan x$, so $\tan y = x$. Differentiate:

$$\sec^2 y \frac{dy}{dx} = 1 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$$

using the identity $\sec^2 y = 1 + \tan^2 y$.

$y = \arctan x$ is steepest at $x = 0$, where its derivative $\tfrac{1}{1 + x^2}$ reaches its maximum of $1$.

The derivatives

Function	Derivative
$\arcsin x$	$\dfrac{1}{\sqrt{1 - x^2}}$
$\arccos x$	$-\dfrac{1}{\sqrt{1 - x^2}}$
$\arctan x$	$\dfrac{1}{1 + x^2}$

The arcsin and arccos formulas hold for $-1 \lt x \lt 1$. At $x = \pm 1$ the denominator is $0$ and the derivative doesn’t exist (the graphs have vertical tangents there). The arctan formula works for every real $x$.

The first three are the ones you must know for AP Calculus AB. For completeness, the other three are (these assume arcsec has range $0 \le y \le \pi$ with $y \ne \tfrac{\pi}{2}$, and arccsc has range $-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}$ with $y \ne 0$; you won’t need them on the AB exam)

$$\frac{d}{dx}\operatorname{arccot} x = -\frac{1}{1 + x^2}, \qquad \frac{d}{dx}\operatorname{arcsec} x = \frac{1}{|x|\sqrt{x^2 - 1}}, \qquad \frac{d}{dx}\operatorname{arccsc} x = -\frac{1}{|x|\sqrt{x^2 - 1}}$$

Notice the “co” functions are just the negatives of their partners.

With the chain rule

If $u$ is a function of $x$:

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

Worked examples

Example 1: Arctan with a constant multiple inside

Differentiate $y = \arctan(3x)$.

Solution. Here $u = 3x$, so $u' = 3$ and $u^2 = 9x^2$:

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

Example 2: Arcsin of a power

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

Solution. Here $u = x^2$, so $u' = 2x$ and $u^2 = x^4$:

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

Example 3: A tangent line

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

Solution. The point: $\arctan 1 = \dfrac{\pi}{4}$, so it’s $\left(1, \dfrac{\pi}{4}\right)$.

The slope: $\dfrac{dy}{dx} = \dfrac{1}{1 + 1^2} = \dfrac{1}{2}$.

$$y - \frac{\pi}{4} = \frac{1}{2}(x - 1)$$

Example 4: Product rule with arcsin

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

Solution. Product rule:

$$f'(x) = 1 \cdot \arcsin x + x \cdot \frac{1}{\sqrt{1 - x^2}}$$

At $x = \dfrac{1}{2}$: $\arcsin\dfrac{1}{2} = \dfrac{\pi}{6}$ and $\sqrt{1 - \dfrac{1}{4}} = \dfrac{\sqrt{3}}{2}$, so

$$f'\left(\frac{1}{2}\right) = \frac{\pi}{6} + \frac{1/2}{\sqrt{3}/2} = \frac{\pi}{6} + \frac{1}{\sqrt{3}} = \frac{\pi}{6} + \frac{\sqrt{3}}{3} \approx 1.101$$

Common mistakes

Reading sin⁻¹ x as 1/sin x. $\sin^{-1} x$ is the inverse sine (arcsine). The reciprocal $\dfrac{1}{\sin x}$ is written $(\sin x)^{-1}$ or $\csc x$, and its derivative is completely different.

Forgetting the chain rule. $\dfrac{d}{dx}\arctan(3x) = \dfrac{3}{1 + 9x^2}$, not $\dfrac{1}{1 + 9x^2}$. The inside derivative goes in the numerator.

Squaring the inside incorrectly. In $\arcsin(x^2)$, $u^2 = x^4$, not $x^2$. In $\arctan(3x)$, $u^2 = 9x^2$, not $3x^2$. Put brackets around $u$ before squaring.

Dropping the negative sign for arccos. $\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1 - x^2}}$. Arccos is a decreasing function, so its slope must be negative.

Using degrees. $\arctan 1 = \dfrac{\pi}{4}$, not $45$, when you’re writing a tangent line or doing any calculus. The derivative formulas only work in radians.

Mixing up the sine and tangent formulas. Arcsin and arccos have a square root, $\sqrt{1 - x^2}$. Arctan has no square root, and a plus sign: $1 + x^2$.

Practice

1. (Warm-up) Evaluate each derivative.

• (a) $\dfrac{d}{dx}\arccos x$ at $x = 0$
• (b) $\dfrac{d}{dx}\arctan x$ at $x = \sqrt{3}$

Solution

(a) $-\dfrac{1}{\sqrt{1 - 0}} = -1$

(b) $\dfrac{1}{1 + (\sqrt{3})^2} = \dfrac{1}{4}$

2. (Warm-up) Differentiate $y = \sin^{-1}(5x)$.

Solution

This is $\arcsin(5x)$, with $u = 5x$ and $u' = 5$:

$$\frac{dy}{dx} = \frac{5}{\sqrt{1 - 25x^2}}$$

3. (Warm-up) Differentiate $y = \arctan\left(\dfrac{x}{2}\right)$ and simplify.

Solution

$u = \dfrac{x}{2}$, $u' = \dfrac{1}{2}$:

$$\frac{dy}{dx} = \frac{1/2}{1 + \dfrac{x^2}{4}}$$

Multiply the top and bottom by $4$:

$$\frac{dy}{dx} = \frac{2}{4 + x^2}$$

4. (Core) Differentiate $y = \arctan(e^x)$.

Solution

$u = e^x$, $u' = e^x$, $u^2 = e^{2x}$:

$$\frac{dy}{dx} = \frac{e^x}{1 + e^{2x}}$$

5. (Core) A camera on the ground is $30$ m from a rocket’s launch pad. When the rocket is $h$ metres high, the camera’s angle of elevation is $\theta = \arctan\left(\dfrac{h}{30}\right)$ radians. Find $\dfrac{d\theta}{dh}$ when $h = 40$, and explain what it means.

Solution

$$\frac{d\theta}{dh} = \frac{1/30}{1 + \left(\dfrac{h}{30}\right)^2}$$

At $h = 40$, $\left(\dfrac{40}{30}\right)^2 = \dfrac{16}{9}$, so

$$\frac{d\theta}{dh} = \frac{1/30}{25/9} = \frac{9}{750} = \frac{3}{250} = 0.012$$

When the rocket is $40$ m high, the angle increases by about $0.012$ radians for each additional metre the rocket climbs.

6. (Core) Differentiate $y = x^2\arccos x$.

Solution

Product rule:

$$\frac{dy}{dx} = 2x\arccos x + x^2 \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right) = 2x\arccos x - \frac{x^2}{\sqrt{1 - x^2}}$$

7. (Core) Find the equation of the tangent line to $y = \arcsin x$ at $x = \dfrac{1}{2}$.

Solution

The point is $\left(\dfrac{1}{2}, \dfrac{\pi}{6}\right)$. The slope is

$$\frac{1}{\sqrt{1 - \frac{1}{4}}} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$$$y - \frac{\pi}{6} = \frac{2\sqrt{3}}{3}\left(x - \frac{1}{2}\right)$$

8. (Challenge) Show that $\dfrac{d}{dx}\big(\arcsin x + \arccos x\big) = 0$ for $-1 \lt x \lt 1$. What does this tell you about $\arcsin x + \arccos x$? Find its value.

Solution

$$\frac{d}{dx}\big(\arcsin x + \arccos x\big) = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - x^2}} = 0$$

A function whose derivative is always $0$ on an interval is constant there. To find the constant, use any convenient $x$, like $x = 0$:

$$\arcsin 0 + \arccos 0 = 0 + \frac{\pi}{2} = \frac{\pi}{2}$$

So $\arcsin x + \arccos x = \dfrac{\pi}{2}$. (It also holds at the endpoints: for example, $\arcsin 1 + \arccos 1 = \dfrac{\pi}{2} + 0$.)

9. (Challenge) Let $g(x) = \arctan\left(\dfrac{1}{x}\right)$ for $x \gt 0$. Find $g'(x)$ and simplify. How does it compare with the derivative of $\arctan x$?

Solution

$u = \dfrac{1}{x} = x^{-1}$, so $u' = -\dfrac{1}{x^2}$:

$$g'(x) = \frac{-1/x^2}{1 + \dfrac{1}{x^2}}$$

Multiply the top and bottom by $x^2$:

$$g'(x) = \frac{-1}{x^2 + 1}$$

This is exactly the negative of $\dfrac{d}{dx}\arctan x$. So $\arctan x + \arctan\left(\dfrac{1}{x}\right)$ has derivative $0$ and is constant for $x \gt 0$. At $x = 1$ it equals $\dfrac{\pi}{4} + \dfrac{\pi}{4} = \dfrac{\pi}{2}$.