Question

find the third order of sin^-1x

Answer 1

   Sin ⁻¹ x

Is this a problem of finding third order term in the Taylor series expansion ?

$\frac{d}{dx}Sin^{-1} x= \frac{1}{\sqrt{1-x^2}}\\\\\frac{d^2}{dx^2}Sin^{-1}x=\frac{-2x}{(1-x^2)^{\frac{3}{2}}}\\\\\frac{d^3}{dx^3}Sin^{-1}x=\frac{-2}{(1-x^2)^{\frac{3}{2}}}-[2x*\frac{-3}{2}\frac{-2x}{(1-x^2))^{\frac{5}{2}}}]\\\\=\frac{-2(1-x^2)-6x^2}{(1-x^2)^{\frac{5}{2}}}\\\\=\frac{-2+x^2-6x^2}{(1-x^2)^{\frac{5}{2}}}\\\\=-\frac{2+5x^2}{(1-x^2)^{\frac{5}{2}}}$

$If\ Sin\ x\ is\ expanded\ at\ x=0\\\\Sin^{-1}x=Sin^{-1}0+\frac{[sin^{-1}x]'|_0}{1!}x+\frac{1}{2!}[sin^{-1}x]''_0x^2+\frac{1}{3!}[sin^{-1}x]'''_0x^3+....\\\\Third\ order\ term\ is:\ \frac{-2}{3!}x^3=-\frac{x^3}{3}$