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Question Number 152918 by ZiYangLee last updated on 03/Sep/21

Find the first derivative of   y=x(√(16−x^2 ))+16sin^(−1) (x/4)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{first}\:\mathrm{derivative}\:\mathrm{of}\: \\ $$$${y}={x}\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }+\mathrm{16sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{4}} \\ $$

Answered by puissant last updated on 03/Sep/21

y=x(√(16−x^2 ))+16arcsin((x/4))  (dy/dx)=(√(16−x^2 ))−((2x^2 )/(2(√(16−x^2 ))))+16((1/4)/( (√(1−(x^2 /(16))))))  =(√(16−x^2 ))−((x^2 (√(16−x^2 )))/(16−x^2 ))+((16(√(16−x^2 )))/(16−x^2 ))  =(√(16−x^2 ))+(√(16−x^2 ))(((16−x^2 )/(16−x^2 )))  =2(√(16−x^2 ))..

$${y}={x}\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }+\mathrm{16}{arcsin}\left(\frac{{x}}{\mathrm{4}}\right) \\ $$$$\frac{{dy}}{{dx}}=\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }}+\mathrm{16}\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\:\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{16}}}} \\ $$$$=\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }−\frac{{x}^{\mathrm{2}} \sqrt{\mathrm{16}−{x}^{\mathrm{2}} }}{\mathrm{16}−{x}^{\mathrm{2}} }+\frac{\mathrm{16}\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }}{\mathrm{16}−{x}^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }\left(\frac{\mathrm{16}−{x}^{\mathrm{2}} }{\mathrm{16}−{x}^{\mathrm{2}} }\right) \\ $$$$=\mathrm{2}\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }.. \\ $$

Commented by ZiYangLee last updated on 04/Sep/21

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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