Trigonometry/Derivative of Inverse Functions

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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 seriesEdit

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 uding Taylor's theorem; it will converge for 1<x<2a-1 .