The inverse functions ${\displaystyle \arcsin(x)}$ , etc. have derivatives that are purely algebraic functions.

If ${\displaystyle y=\arcsin(x)}$ then ${\displaystyle x=\sin(y)}$ and

${\displaystyle {\frac {dx}{dy}}=\cos(y)={\sqrt {1-\sin ^{2}(y)}}={\sqrt {1-x^{2}}}}$ .

So

${\displaystyle {\frac {dy}{dx}}={\frac {1}{\frac {dx}{dy}}}={\frac {1}{\sqrt {1-x^{2}}}}}$

Similarly,

${\displaystyle {\frac {d}{dx}}{\bigl [}\arccos(x){\bigr ]}=-{\frac {1}{\sqrt {1-x^{2}}}}}$ .

If ${\displaystyle y=\arctan(x)}$ then ${\displaystyle x=\tan(y)}$ and

${\displaystyle {\frac {dx}{dy}}=\sec ^{2}(y)=1+\tan ^{2}(y)=1+x^{2}}$ .

So

${\displaystyle {\frac {dy}{dx}}={\frac {1}{\frac {dx}{dy}}}={\frac {1}{1+x^{2}}}}$

If ${\displaystyle y=\operatorname {arcsec}(x)}$ then ${\displaystyle x=\sec(y)}$ and

${\displaystyle {\frac {dx}{dy}}=\sec(y)\tan(y)=x{\sqrt {x^{2}-1}}}$ .

So

${\displaystyle {\frac {dy}{dx}}={\frac {1}{\frac {dx}{dy}}}={\frac {1}{x{\sqrt {x^{2}-1}}}}}$

Power series

The above results provide an easy way to find the power series expansions of these functions.

${\displaystyle {\frac {1}{\sqrt {1-x^{2}}}}=1+{\frac {x^{2}}{2}}+{\frac {3x^{4}}{8}}+{\frac {5x^{6}}{16}}+{\frac {35x^{8}}{128}}+\cdots }$

This is uniformly convergent if ${\displaystyle |x|<1}$ so can be integrated term by term. The constant of integration is zero since ${\displaystyle \arcsin(0)=0}$ , so

${\displaystyle \arcsin(x)=x+{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}+{\frac {5x^{7}}{112}}+{\frac {35x^{9}}{1152}}+\cdots }$

${\displaystyle {\frac {1}{1+x^{2}}}=1-x^{2}+x^{4}-x^{6}+\cdots }$

This is uniformly convergent if ${\displaystyle |x|<1}$ so can be integrated term by term. The constant of integration is zero since ${\displaystyle \arctan(0)=0}$ , so

${\displaystyle \arctan(x)=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots }$

Note that ${\displaystyle \operatorname {arcsec}(x)}$ has no power series expansion about ${\displaystyle x=0}$ , as it is not defined for ${\displaystyle x<1}$ and has an infinite derivative when ${\displaystyle x=1}$ . An expansion about any point ${\displaystyle x=a>1}$ in powers of ${\displaystyle x-a}$ can be found uding Taylor's theorem; it will converge for ${\displaystyle 1<x<2a-1}$ .