Trigonometry/Derivative of Inverse Functions

The inverse functions sin^-1(x), etc. have derivatives that are purely algebraic functions.

If y = sin^-1(x) then 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,

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

If y = tan^-1(x) then 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 y = sec^-1(x) then 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.

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

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

$\displaystyle \sin^{-1}(x) = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} ...$

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

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

$\displaystyle \tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ...$

Note that sec^-1(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 uding Taylor's theorem; it will converge for 1 < x < 2a-1.