Derivatives of Inverse Trig Functions

The derivatives of inverse trig functions are algebraic, with no trig left in them. The two you will use most are $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ and $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$ . Arccosine, arccotangent and arccosecant have the same derivatives as their partners, with a minus sign.

Before this lesson: Inverse Trig Functions, Derivatives of Trig Functions and Implicit Differentiation

Where the formulas come from

Start with $y = \arcsin x$ , which means $\sin y = x$ with y between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ . Differentiate both sides with respect to x:

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

Now write cos y in terms of x. From the Pythagorean identity, $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ (see fundamental trig identities). The square root is positive because cos y ≥ 0 for y in that interval. So the derivative of arcsin x is $\frac{1}{\sqrt{1 - x^2}}$ . The same steps with $\tan y = x$ and $\sec^2 y = 1 + \tan^2 y$ give the arctangent formula.

Graph of y = arctan x for x from −6 to 6, between dashed horizontal asymptotes at y = π/2 and y = −π/2. Dashed tangent lines are drawn at (0, 0) with slope 1 and at (1, π/4) with slope 1/2. The curve flattens as x moves away from 0. — The slope of arctan x is 1/(1 + x²): it is 1 at x = 0, 1/2 at x = 1, and close to 0 far from the origin.

Inverse trig derivative formulas

Function	Derivative	Function	Derivative
$\arcsin x$	$\frac{1}{\sqrt{1 - x^2}}$	$\arccos x$	$-\frac{1}{\sqrt{1 - x^2}}$
$\arctan x$	$\frac{1}{1 + x^2}$	$\text{arccot}\, x$	$-\frac{1}{1 + x^2}$
$\text{arcsec}\, x$	$\frac{1}{\|x\|\sqrt{x^2 - 1}}$	$\text{arccsc}\, x$	$-\frac{1}{\|x\|\sqrt{x^2 - 1}}$

The arcsine and arccosine formulas need $-1 \lt x \lt 1$ , and the arcsecant pair needs $|x| \gt 1$ . Arctangent works for every x. You may also see the notation $\sin^{-1} x$ for arcsin x; it means the inverse, not 1/sin x.

With an inside function u, multiply by u′ (chain rule): for example, $\frac{d}{dx}\arctan u = \frac{u'}{1 + u^2}$ .

Derivatives of inverse trig functions examples

Example 1: Arcsine with the chain rule

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

u = 3x, \; u' = 3

The inside function is 3x.

y' = \frac{1}{\sqrt{1 - (3x)^2}} \cdot 3

Arcsine formula with u = 3x, times u′.

y' = \frac{3}{\sqrt{1 - 9x^2}}

Square 3x and multiply.

Answer: $y' = \frac{3}{\sqrt{1 - 9x^2}}$ , for $-\frac{1}{3} \lt x \lt \frac{1}{3}$ .

Example 2: Arctangent with the chain rule

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

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

Arctangent formula with u = x², times u′ = 2x.

y' = \frac{2x}{1 + x^4}

Simplify: (x²)² = x⁴.

Answer: $y' = \frac{2x}{1 + x^4}$ .

Example 3: Product rule with arctangent

Differentiate $y = x\arctan x$ .

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

Product rule (see product rule).

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

Simplify.

Answer: $y' = \arctan x + \frac{x}{1 + x^2}$ .

Example 4: A tangent line to arcsine

Find the tangent line to $y = \arcsin x$ at $x = \frac{1}{2}$ .

y = \arcsin\frac{1}{2} = \frac{\pi}{6}

Find the point. The angle whose sine is ½ is π/6.

y' = \frac{1}{\sqrt{1 - \frac{1}{4}}} = \frac{1}{\sqrt{\frac{3}{4}}}

Substitute x = ½ into the derivative.

y' = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

√(3/4) = √3/2, then rationalize (see rationalizing the denominator).

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

Use point-slope form (see point-slope form).

Answer: $y = \frac{2\sqrt{3}}{3}\left(x - \frac{1}{2}\right) + \frac{\pi}{6}$ , or about y = 1.155x − 0.054.

Example 5: Word problem with a rocket

A rocket rises straight up from a launch pad. An observer stands 300 m from the pad. When the rocket is h meters high, the angle of elevation is $\theta = \arctan\frac{h}{300}$ . How fast is the angle changing per meter of height when h = 400 m? If the rocket climbs at 50 m/s, how fast is the angle changing per second?

\frac{d\theta}{dh} = \frac{1}{1 + \left(\frac{h}{300}\right)^2} \cdot \frac{1}{300}

Arctangent formula with u = h/300 and u′ = 1/300.

\frac{d\theta}{dh} = \frac{300}{300^2 + h^2}

Multiply the top and bottom by 300².

\frac{d\theta}{dh} = \frac{300}{90000 + 160000} = \frac{300}{250000}

Substitute h = 400.

\frac{d\theta}{dh} = 0.0012

Divide. The units are radians per meter.

\frac{d\theta}{dt} = 0.0012 \cdot 50 = 0.06

Chain rule: dθ/dt = (dθ/dh)(dh/dt).

Answer: at 400 m, the angle grows by 0.0012 radian per meter, or 0.06 radian (about 3.4°) per second. This kind of chained rate is the topic of the next lesson, related rates.

Common mistakes

Mistake	Example of the error	Fix
Reading sin⁻¹x as 1/sin x	$\frac{d}{dx}\sin^{-1}x = -\frac{\cos x}{\sin^2 x}$	$\sin^{-1}x$ is arcsin x: the derivative is $\frac{1}{\sqrt{1 - x^2}}$ .
Mixing up the arcsine and arctangent formulas	$\frac{d}{dx}\arctan x = \frac{1}{\sqrt{1 - x^2}}$	Arctangent has no root: $\frac{1}{1 + x^2}$ .
Forgetting to square the inside	$\frac{d}{dx}\arcsin(3x) = \frac{3}{\sqrt{1 - 3x^2}}$	Square all of u: $(3x)^2 = 9x^2$ .
Forgetting the chain rule factor	$\frac{d}{dx}\arctan(x^2) = \frac{1}{1 + x^4}$	Multiply by u′ = 2x.
Dropping the minus sign for the co-functions	$\frac{d}{dx}\arccos x = \frac{1}{\sqrt{1 - x^2}}$	Arccosine falls, so its derivative is negative.

Related methods

Using the inverse function rule

You can also get these formulas from the rule for derivatives of inverse functions: $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ . With f = sin, that is $\frac{1}{\cos(\arcsin x)} = \frac{1}{\sqrt{1 - x^2}}$ .

Why these derivatives matter for integrals

Read backward, the formulas say that an antiderivative of $\frac{1}{1 + x^2}$ is arctan x and an antiderivative of $\frac{1}{\sqrt{1 - x^2}}$ is arcsin x. Integrals like these show up later in Calculus 1, in Integrals That Give Inverse Trig Functions.

Practice problems

Differentiate $y = \arcsin x + \arccos x$ . What does the answer tell you?
Differentiate $y = \arctan(5x)$ .
Differentiate $y = \arcsin\frac{x}{2}$ .
Find the slope of $y = \text{arcsec}\, x$ at x = 2.
Differentiate $y = \arctan\sqrt{x}$ .
Differentiate $y = (\arcsin x)^2$ .
Differentiate $y = \arctan(e^x)$ .
Find the tangent line to $y = \arctan x$ at x = 1.
Differentiate $y = x^2\arccos x$ .
Differentiate $y = \arctan\frac{1}{x}$ and simplify.

Show answers

y' = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - x^2}}

Use both formulas.

y' = 0

The terms cancel.

Answer: y′ = 0, so arcsin x + arccos x is a constant. (It equals π/2.)

y' = \frac{1}{1 + (5x)^2} \cdot 5

Arctangent formula with u = 5x.

y' = \frac{5}{1 + 25x^2}

Simplify.

Answer: $y' = \frac{5}{1 + 25x^2}$

y' = \frac{1}{\sqrt{1 - \frac{x^2}{4}}} \cdot \frac{1}{2}

Arcsine formula with u = x/2 and u′ = ½.

y' = \frac{1}{2\sqrt{\frac{4 - x^2}{4}}} = \frac{1}{\sqrt{4 - x^2}}

Write 1 − x²/4 over 4; √4 = 2 cancels the 2.

Answer: $y' = \frac{1}{\sqrt{4 - x^2}}$

y' = \frac{1}{|x|\sqrt{x^2 - 1}}

Arcsecant formula.

y'(2) = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6}

Substitute x = 2 and rationalize.

Answer: $\frac{\sqrt{3}}{6} \approx 0.289$

y' = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}}

Arctangent formula with u = √x; (√x)² = x and u′ = 1/(2√x).

y' = \frac{1}{2\sqrt{x}(1 + x)}

Multiply.

Answer: $y' = \frac{1}{2\sqrt{x}(1 + x)}$

y' = 2\arcsin x \cdot \frac{1}{\sqrt{1 - x^2}}

General power rule: the inside is arcsin x.

Answer: $y' = \frac{2\arcsin x}{\sqrt{1 - x^2}}$

y' = \frac{1}{1 + (e^x)^2} \cdot e^x

Arctangent formula with u = eˣ.

y' = \frac{e^x}{1 + e^{2x}}

(eˣ)² = e²ˣ.

Answer: $y' = \frac{e^x}{1 + e^{2x}}$

y(1) = \arctan 1 = \frac{\pi}{4}

Find the point.

y'(1) = \frac{1}{1 + 1} = \frac{1}{2}

Substitute x = 1 into 1/(1 + x²).

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

Point-slope form.

Answer: $y = \frac{1}{2}x - \frac{1}{2} + \frac{\pi}{4}$ , or about y = 0.5x + 0.285

y' = 2x\arccos x + x^2 \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right)

Product rule.

Answer: $y' = 2x\arccos x - \frac{x^2}{\sqrt{1 - x^2}}$

10.

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

Arctangent formula with u = 1/x and u′ = −1/x².

y' = \frac{-1}{x^2 + 1}

Multiply the denominators: x²(1 + 1/x²) = x² + 1.

Answer: $y' = -\frac{1}{x^2 + 1}$

Derivatives of inverse trig functions FAQ

What is the derivative of arctan x?

$\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$ , for every real x.

What is the derivative of arcsin x?

$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ , for $-1 \lt x \lt 1$ . At x = ±1 the graph has vertical tangents, so there is no derivative.

Is sin⁻¹x the same as 1/sin x?

No. $\sin^{-1}x$ means arcsin x, the inverse function. The reciprocal 1/sin x is csc x.

Why is the derivative of arccos x negative?

Arccosine decreases from π to 0 as x goes from −1 to 1, so its slope is always negative. Its derivative is the arcsine derivative with a minus sign.

Summary

$(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}$ and $(\arccos x)' = -\frac{1}{\sqrt{1 - x^2}}$ .
$(\arctan x)' = \frac{1}{1 + x^2}$ and $(\text{arcsec}\, x)' = \frac{1}{|x|\sqrt{x^2 - 1}}$ .
The co-functions get a minus sign.
With an inside u, multiply by u′ and square all of u.
The formulas come from implicit differentiation and the Pythagorean identity.

Previous lesson: Logarithmic Differentiation

Next lesson: Related Rates

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