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