Derivatives of the Inverse Trigonometric Functions

Function	Derivative	Domain
$\arcsin x$	${\displaystyle \frac{1}{\sqrt{1-x^2}}}$
$\arccos x$	$-{\displaystyle \frac{1}{\sqrt{1-x^2}}}$
$\arctan x$	${\displaystyle \frac{1}{1+x^2}}$	all $x$

Recall: Inverse Sine

$$v=\arcsin u={\sin }^{-1}u\text{means}u=\sin v\text{and}-\frac{\pi }{2}\le v\le \frac{\pi }{2}$$

Note that $\arcsin x$ and ${\sin }^{-1}x$ denote the same thing: the inverse of the sine function with domain restricted to $[-\pi /2,\pi /2]$.

Derivative of $\arccos x$

$$v=\arccos u\text{means}u=\cos v\text{and}0\le v\le \pi$$

Derivative of $\arctan x$

$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$$

Combined Chain-Rule Examples

When the argument of an inverse trig function is a composite expression $u(x)$, apply the Chain Rule:

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

Frequently Asked Questions

Why are the derivatives of arcsin and arccos negatives of each other?

Because $\arcsin x+\arccos x=\pi /2$ for all $x\in [-1,1]$, differentiating both sides gives (\arcsin x)'+(\arccos x)'=0, so the two derivatives sum to zero. This confirms (\arccos x)'=-(\arcsin x)'.

Where is $\arctan x$ differentiable?

$\arctan x$ is differentiable for all real $x$, with derivative . The function is increasing everywhere and has horizontal asymptotes at $y=\pm \pi /2$.

What is ${\displaystyle \frac{d}{dx}}\arctan (3x^2)$?

By the Chain Rule: ${\displaystyle \frac{1}{1+(3x^2{)}^2}}\cdot 6x={\displaystyle \frac{6x}{1+9x^4}}$.