Calculus Reference

Inverse Trigonometric Derivatives

Inverse trigonometric functions return angles. Their derivatives combine a standard formula with the chain rule when the input is more than just $x$.

Solved Examples

Learn the basic arctangent rule, then apply inverse sine and cosine to composite expressions.

Differentiate arctangent
Differentiate inverse sine with the chain rule
Differentiate inverse cosine with the chain rule

Inverse Trigonometric Derivative Table

Function	Derivative	Input condition
$\arcsin(u)$	$\frac{u'}{\sqrt{1-u^2}}$	$\|u\|<1$ for a finite derivative.
$\arccos(u)$	$-\frac{u'}{\sqrt{1-u^2}}$	$\|u\|<1$ for a finite derivative.
$\arctan(u)$	$\frac{u'}{1+u^2}$	All real $u$.
$\operatorname{arccot}(u)$	$-\frac{u'}{1+u^2}$	Standard range convention $(0,\pi)$.
$\operatorname{arcsec}(u)$	$\frac{u'}{\|u\|\sqrt{u^2-1}}$	$\|u\|>1$ for a finite derivative.
$\operatorname{arccsc}(u)$	$-\frac{u'}{\|u\|\sqrt{u^2-1}}$	$\|u\|>1$ for a finite derivative.

The vertical bars in the arcsecant and arccosecant formulas matter when the input is negative. At endpoint inputs, arcsine and arccosine are defined but their derivative formulas are not finite.

Why the Chain Rule Appears

Each formula assumes the inverse trigonometric function receives an input $u(x)$. Differentiate the outside inverse function, keep $u$, and multiply by $u'(x)$. For example, $\frac{d}{dx}[\arctan(u)]=\frac{u'}{1+u^2}$.

Inverse trigonometric notation such as $\arcsin x$ means the inverse function. It does not mean $1/\sin x$; that reciprocal is $\csc x$. See trigonometric derivatives for sine, cosine, tangent, and their reciprocals.

Example: Basic Arctangent Derivative

Differentiate $f(x)=\arctan x$

For the basic function, the inside is $u=x$ and $u'=1$.

Solution Steps

Use $u'/(1+u^2)$ with $u=x$.
Simplify the denominator.

$$f'(x)=\frac{1}{1+x^2}$$$$\boxed{f'(x)=\frac{1}{1+x^2}}$$

Example: Inverse Sine with the Chain Rule

Differentiate $g(x)=\arcsin(x/2)$

The input must satisfy $-2<x<2$ for a finite derivative.

Solution Steps

Use $u'/\sqrt{1-u^2}$ with $u=x/2$.
Simplify the square root and state the interval.

$$g'(x)=\frac{1/2}{\sqrt{1-(x/2)^2}}$$$$\boxed{g'(x)=\frac{1}{\sqrt{4-x^2}}},\quad -2<x<2$$

Example: Inverse Cosine with the Chain Rule

Differentiate $h(x)=\arccos(2x-1)$

The input is between $-1$ and $1$ when $0<x<1$.

Solution Steps

Use $-u'/\sqrt{1-u^2}$ with $u=2x-1$.
Substitute $u'=2$ and preserve the interval where the derivative is finite.

$$h'(x)=-\frac{2}{\sqrt{1-(2x-1)^2}}$$$$\boxed{h'(x)=-\frac{2}{\sqrt{1-(2x-1)^2}}},\quad 0<x<1$$

Continue Studying Derivatives

Review the chain rule, the special-function guide, and the trigonometric derivative table.