Trigonometry/Derivative of Inverse Functions

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

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

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

So

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

Similarly,

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

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

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

So

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

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

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

So

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

Power series

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

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

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

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

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

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

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

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