Question Number 20205 by vivek last updated on 24/Aug/17
$${find}\:{the}\:{sin}^{−\mathrm{1}} \:{diferentiation} \\ $$
Answered by Joel577 last updated on 24/Aug/17
$${y}\:=\:\mathrm{sin}^{−\mathrm{1}} \:\left({x}\right) \\ $$$${x}\:=\:\mathrm{sin}\:{y} \\ $$$$\frac{{dx}}{{dy}}\:=\:\mathrm{cos}\:{y}\:=\:\sqrt{\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \:{y}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{cos}\:{y}}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \:{y}}}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}\: \\ $$