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Question Number 44267 by rahul 19 last updated on 25/Sep/18

If xε ((1/(√2)) , 1) ,differentiate cos^(−1) (2x(√(1−x^2 ))).

$${If}\:{x}\epsilon\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:,\:\mathrm{1}\right)\:,{differentiate}\:\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right). \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18

x=cosα dx=−sinα dα y=cos^(−1) (2cosαsinα) y=cos^(−1) (sin2α) now xε((1/(√2)),1) α ε (0,(Π/4)) so 2α ε (0,(Π/2)) when x=(1/(√2)) α=(Π/4) x=1 α=0 y=cos^(−1) {cos((Π/2)−2α)} y=(Π/2)−2α (dy/dα)=−2 (dy/dx)=(dy/dα)×(dα/dx)=−2×((−1)/(sinα))=(2/((√(1−x^2 )) )) or approach let x=sinα dx=cosαdα y=cos^(−1) (2sinαcosα) y=cos^(−1) (sin2α) xε((1/(√2)),1) α ε((Π/4),(Π/2)) but 2α ε((Π/2),Π) let 2α=(Π/2)+β 2dα=dβ y=cos^− {sin((Π/2)+β)} y=cos^(−1) {cosβ) y=β dy=dβ (dy/dx)=(dy/dβ)×(dβ/dα)×(dα/dx)=1×2×(1/(cosα))=(2/((√(1−x^2 )) )) pls check

$${x}={cos}\alpha\:\:\:{dx}=−{sin}\alpha\:{d}\alpha \\ $$$${y}={cos}^{−\mathrm{1}} \left(\mathrm{2}{cos}\alpha{sin}\alpha\right) \\ $$$${y}={cos}^{−\mathrm{1}} \left({sin}\mathrm{2}\alpha\right) \\ $$$${now}\:{x}\epsilon\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}},\mathrm{1}\right)\:\:\alpha\:\epsilon\:\left(\mathrm{0},\frac{\Pi}{\mathrm{4}}\right)\:\:\:{so}\:\mathrm{2}\alpha\:\epsilon\:\left(\mathrm{0},\frac{\Pi}{\mathrm{2}}\right) \\ $$$${when}\:{x}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\:\:\alpha=\frac{\Pi}{\mathrm{4}} \\ $$$$\:\:\:\:{x}=\mathrm{1}\:\:\:\:\alpha=\mathrm{0} \\ $$$${y}={cos}^{−\mathrm{1}} \left\{{cos}\left(\frac{\Pi}{\mathrm{2}}−\mathrm{2}\alpha\right)\right\} \\ $$$${y}=\frac{\Pi}{\mathrm{2}}−\mathrm{2}\alpha \\ $$$$\frac{{dy}}{{d}\alpha}=−\mathrm{2}\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{{dy}}{{d}\alpha}×\frac{{d}\alpha}{{dx}}=−\mathrm{2}×\frac{−\mathrm{1}}{{sin}\alpha}=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:} \\ $$$$\:{or}\:{approach} \\ $$$${let}\:{x}={sin}\alpha \\ $$$${dx}={cos}\alpha{d}\alpha \\ $$$${y}={cos}^{−\mathrm{1}} \left(\mathrm{2}{sin}\alpha{cos}\alpha\right) \\ $$$${y}={cos}^{−\mathrm{1}} \left({sin}\mathrm{2}\alpha\right) \\ $$$${x}\epsilon\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}},\mathrm{1}\right)\:\:\:\alpha\:\epsilon\left(\frac{\Pi}{\mathrm{4}},\frac{\Pi}{\mathrm{2}}\right)\:\:{but}\:\mathrm{2}\alpha\:\epsilon\left(\frac{\Pi}{\mathrm{2}},\Pi\right) \\ $$$${let}\:\mathrm{2}\alpha=\frac{\Pi}{\mathrm{2}}+\beta\:\:\:\mathrm{2}{d}\alpha={d}\beta \\ $$$${y}={cos}^{−} \left\{{sin}\left(\frac{\Pi}{\mathrm{2}}+\beta\right)\right\} \\ $$$${y}={cos}^{−\mathrm{1}} \left\{{cos}\beta\right)\:\:\:{y}=\beta \\ $$$${dy}={d}\beta \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{d}\beta}×\frac{{d}\beta}{{d}\alpha}×\frac{{d}\alpha}{{dx}}=\mathrm{1}×\mathrm{2}×\frac{\mathrm{1}}{{cos}\alpha}=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}} \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18

most welcome...you are a good bowler...

$${most}\:{welcome}...{you}\:{are}\:{a}\:{good}\:{bowler}... \\ $$

Commented by rahul 19 last updated on 25/Sep/18

thanks sir ! ����

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