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Question-43438

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Question Number 43438 by peter frank last updated on 10/Sep/18
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
cosh^− (3x)=ln(3x+(√(9x^2 −1)) ) y={ln(3x+(√(9x^2 −1)) )}^2 (dy/dx)=2{ln(3x+(√(9x^2 −1)) )}×(1/(3x+(√(9x^2 −1)) ))×(3+(1/2)×((18x)/( (√(9x^2 −1)) ))) (dy/dx)=2(√y) ×(1/(3x+(√(9x^2 −1)) ))×(3+((9x)/( (√(9x^2 −1))))) (dy/dx)=2(√y) ×(1/(3x+(√(9x^2 −1)) ))×3(1+((3x)/( (√(9x^2 −1))))) (dy/dx)=2(√y) ×(3/( (√(9x^2 −1)))) (√(9x^2 −1)) ×(dy/dx)=6(√y) (9x^2 −1)×((dy/dx))^2 =36y proved...
$${cosh}^{−} \left(\mathrm{3}{x}\right)={ln}\left(\mathrm{3}{x}+\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:\right) \\ $$$${y}=\left\{{ln}\left(\mathrm{3}{x}+\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:\right)\right\}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\left\{{ln}\left(\mathrm{3}{x}+\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:\right)\right\}×\frac{\mathrm{1}}{\mathrm{3}{x}+\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:}×\left(\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{18}{x}}{\:\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:}\right) \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\sqrt{{y}}\:×\frac{\mathrm{1}}{\mathrm{3}{x}+\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:}×\left(\mathrm{3}+\frac{\mathrm{9}{x}}{\:\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}}\right) \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\sqrt{{y}}\:×\frac{\mathrm{1}}{\mathrm{3}{x}+\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:}×\mathrm{3}\left(\mathrm{1}+\frac{\mathrm{3}{x}}{\:\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}}\right) \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\sqrt{{y}}\:×\frac{\mathrm{3}}{\:\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:×\frac{{dy}}{{dx}}=\mathrm{6}\sqrt{{y}}\: \\ $$$$\left(\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}\right)×\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\mathrm{36}{y}\:{proved}… \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
2)(9x^2 −1)((dy/dx))^2 =36y (9x^2 −1)×2((dy/dx))×(d^2 y/dx^2 )+((dy/dx))^2 ×(18x)=36(dy/dx) (9x^2 −1)×(d^2 y/dx^2 )+9x((dy/dx))=18...proved
$$\left.\mathrm{2}\right)\left(\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\mathrm{36}{y} \\ $$$$\left(\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}\right)×\mathrm{2}\left(\frac{{dy}}{{dx}}\right)×\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} ×\left(\mathrm{18}{x}\right)=\mathrm{36}\frac{{dy}}{{dx}} \\ $$$$\left(\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}\right)×\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{9}{x}\left(\frac{{dy}}{{dx}}\right)=\mathrm{18}…{proved} \\ $$

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