# Derivatives of Inverse Trig Functions

Sub Topics
Since the trigonometric functions are periodic functions, the function values repeat many times in their respective domains. You can observe the graphs of trigonometric functions fail the horizontal line test for inverse functions. But the inverse of trigonometric functions can be defined by suitably restricting their domains.  The following table lists the inverse trigonometric functions along with their derivatives.

## Inverse Trig Functions Table

 Function and its domain Derivative and its domain arcsin(x)        -1$\leq$x$\leq$1 $\frac{1}{\sqrt{1-x^{2}}}$             -1 < x < 1 arccos(x)         -1$\leq$x$\leq$1 -$\frac{1}{\sqrt{1-x^{2}}}$            -1 < x < 1 arctan(x)         (-$\infty$,$\infty$) $\frac{1}{1+x^{2}}$                (-$\infty$,$\infty$) arccsc(x) ) (-$\infty$,-1] U [1,$\infty$) $\frac{-1}{x\sqrt{x^{2}-1}}$ (-$\infty$,-1] U [1,$\infty$) arccsc(x)   (-$\infty$,-1] U [1,$\infty$) $\frac{-1}{x\sqrt{x^{2}-1}}$ (-$\infty$,-1] U [1,$\infty$) arccot(x)       (-$\infty$,$\infty$) $\frac{-1}{1+x^{2}}$                 (-$\infty$,$\infty$)

 Derivatives for composite inverse trigonometric functions If u is a differentiable function of x and u' =$\frac{du}{dx}$ $\frac{d}{dx}$ (arcsin u) = $\frac{u'}{\sqrt{1-u^{2}}}$ $\frac{d}{dx}$ (arccos u) = -$\frac{u'}{\sqrt{1-u^{2}}}$ $\frac{d}{dx}$ (arctan u) = $\frac{u'}{1+u^{2}}$ $\frac{d}{dx}$ (arccot u) = -$\frac{u'}{1+u^{2}}$ $\frac{d}{dx}$ (arcsec u) = $\frac{u'}{|u|\sqrt{u^{2}-1}}$ $\frac{d}{dx}$ (arccsc u) = $\frac{-u'}{|u|\sqrt{x^{2}-1}}$

## Differentiating Inverse Trigonometric Functions

Inverse trigonometric functions can often be simplified using some trigonometric substitution before finding the derivative.  The simplified function can be differentiated using chain rule to obtain derivative of the given function. The double angle and half angle formulas for trigonometric functions hint at the substitution to be used.