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,

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

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 $y=\mathrm{arcsec}(x)$ 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 $\mathrm{arcsec}(x)$ 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 $1<x<2a-1$ .