if-y-sin-1-2x-1-x-2-where-x-1-2-1-2-then-find-dy-dx-

Question Number 190660 by vishal1234 last updated on 08/Apr/23

$${if}\:{y}\:=\:{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\:{where}\:{x}\in\left[−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right],\:{then}\:{find}\:\frac{{dy}}{{dx}}. \\ $$

Answered by BaliramKumar last updated on 08/Apr/23

$$\mathrm{let},\:{x}\:=\:{sin}\theta\:\Rightarrow\:\theta\:=\:{sin}^{−\mathrm{1}} {x} \\ $$$${y}\:=\:{sin}^{−\mathrm{1}} \left(\mathrm{2}{sin}\theta\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta}\right) \\ $$$${y}\:=\:{sin}^{−\mathrm{1}} \left(\mathrm{2}{sin}\theta\sqrt{{cos}^{\mathrm{2}} \theta}\right) \\ $$$${y}\:=\:{sin}^{−\mathrm{1}} \left(\mathrm{2}{sin}\theta\centerdot{cos}\theta\right) \\ $$$${y}\:=\:{sin}^{−\mathrm{1}} \left({sin}\mathrm{2}\theta\right) \\ $$$${y}\:=\:\mathrm{2}\theta\:=\:\mathrm{2}{sin}^{−\mathrm{1}} {x} \\ $$$$\frac{{dy}}{{dx}}\:=\:\mathrm{2}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\: \\ $$$$ \\ $$$$ \\ $$

Commented by vishal1234 last updated on 14/Apr/23

$${what}\:{will}\:{happen},?\:{if}\:{we}\:{take}\:{x}\:=\:{cos}\theta? \\ $$

Answered by horsebrand11 last updated on 09/Apr/23

$$\:{y}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right) \\ $$$$\:\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:=\:\mathrm{sin}\:{y} \\ $$$$\:\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\mathrm{2}{x}\left(\frac{−\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)=\:\mathrm{cos}\:{y}\:.\frac{{dy}}{{dx}} \\ $$$$\:\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\:\mathrm{cos}\:{y}.\:\frac{{dy}}{{dx}} \\ $$$$\:\frac{\mathrm{2}−\mathrm{4}{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)}}\:=\:\frac{{dy}}{{dx}} \\ $$$$\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}−\mathrm{4}{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\sqrt{\mathrm{4}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}}\: \\ $$$$\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}−\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\: \\ $$

Commented by som(math1967) last updated on 09/Apr/23

$$\mathrm{4}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}=\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:{or}\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$

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