Derivatives of Inverse Trigonometric Functions

By the end of this lesson, you will be able to differentiate all six inverse trig functions and combine them with the Chain Rule.

The Inverse Trig Derivatives

$$\frac{d}{dx}[\sin^{-1}x] = \frac{1}{\sqrt{1-x^2}}$$

$$\frac{d}{dx}[\cos^{-1}x] = -\frac{1}{\sqrt{1-x^2}}$$

$$\frac{d}{dx}[\tan^{-1}x] = \frac{1}{1+x^2}$$

$$\frac{d}{dx}[\cot^{-1}x] = -\frac{1}{1+x^2}$$

$$\frac{d}{dx}[\sec^{-1}x] = \frac{1}{|x|\sqrt{x^2-1}}$$

$$\frac{d}{dx}[\csc^{-1}x] = -\frac{1}{|x|\sqrt{x^2-1}}$$

Pattern: co-inverse functions (arccos, arccot, arccsc) have the same formula with a negative sign.

Deriving $(\sin^{-1}x)' = 1/\sqrt{1-x^2}$

Let $y = \sin^{-1}x$, so $\sin y = x$ with $y \in [-\pi/2, \pi/2]$.

Differentiate implicitly: $\cos y \cdot y' = 1 \implies y' = \dfrac{1}{\cos y}$

Since $\cos^2 y = 1 - \sin^2 y = 1 - x^2$ and $\cos y > 0$ (as $y \in [-\pi/2, \pi/2]$):

$y' = \dfrac{1}{\sqrt{1-x^2}}$ ✓

Deriving $(\tan^{-1}x)' = 1/(1+x^2)$

Let $y = \tan^{-1}x$, so $\tan y = x$.

$\sec^2 y \cdot y' = 1 \implies y' = \cos^2 y = \dfrac{1}{\sec^2 y} = \dfrac{1}{1+\tan^2 y} = \dfrac{1}{1+x^2}$ ✓

Example 1 — Direct Application

$\dfrac{d}{dx}[\sin^{-1}(3x)]$

$= \dfrac{1}{\sqrt{1-9x^2}} \cdot 3 = \dfrac{3}{\sqrt{1-9x^2}}$ (Chain Rule)

Example 2 — Combination with Product Rule

$f(x) = x^2\tan^{-1}(x)$

$f'(x) = 2x\tan^{-1}x + x^2 \cdot \dfrac{1}{1+x^2} = 2x\tan^{-1}x + \dfrac{x^2}{1+x^2}$

Example 3 — Chain Rule with Arccos

$g(x) = \cos^{-1}(e^x)$

$g'(x) = -\dfrac{1}{\sqrt{1-e^{2x}}} \cdot e^x = -\dfrac{e^x}{\sqrt{1-e^{2x}}}$

Domain requires $|e^x| \leq 1 \implies x \leq 0$.

Common Antiderivative Patterns

The inverse trig derivatives appear as antiderivatives in integration:

$$\int\frac{1}{\sqrt{1-x^2}},dx = \sin^{-1}x + C$$

$$\int\frac{1}{1+x^2},dx = \tan^{-1}x + C$$

Summary

Function	Derivative
$\sin^{-1}x$	$1/\sqrt{1-x^2}$