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Question Number 20832 by j.masanja06@gmail.com last updated on 04/Sep/17

Given that y=log(√((1−cos^2 x)/(1−e^(2x) ))) find (dy/dx)

$${Given}\:{that}\:{y}={log}\sqrt{\frac{\mathrm{1}−{cos}^{\mathrm{2}} {x}}{\mathrm{1}−{e}^{\mathrm{2}{x}} }} \\ $$$${find}\:\frac{{dy}}{{dx}} \\ $$

Commented by myintkhaing last updated on 04/Sep/17

ecos^2 x ???

$$\mathrm{ecos}^{\mathrm{2}} \mathrm{x}\:??? \\ $$

Commented by geovane10math last updated on 05/Sep/17

This “log” is the natural logarithm? ln?

$$\mathrm{This}\:``\mathrm{log}''\:\mathrm{is}\:\mathrm{the}\:\mathrm{natural}\:\mathrm{logarithm}? \\ $$$$\mathrm{ln}? \\ $$

Answered by Hitler last updated on 05/Sep/17

((sgn(sinx)cosx+(∣sinx∣−sgn(sinx)cosx)e^(2x) )/((1−e^(2x) )∣sinx∣ln10))

$$\frac{{sgn}\left({sinx}\right){cosx}+\left(\mid{sinx}\mid−{sgn}\left({sinx}\right){cosx}\right){e}^{\mathrm{2}{x}} }{\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right)\mid{sinx}\mid{ln}\mathrm{10}} \\ $$

Answered by myintkhaing last updated on 05/Sep/17

y=log(√((sin^2 x)/(1−e^(2x) )))=log((sin x)/(√(1−e^(2x) ))) y′=((√(1−e^(2x) ))/(sin x)) (((√(1−e^(2x) ))cos x−sin x(1/(2(√(1−e^(2x) ))))(−2e^(2x) ))/(1−e^(2x) )) =((√(1−e^(2x) ))/(sin x)) (((1−e^(2x) )cos x+e^(2x) sin x)/((1−e^(2x) )(√(1−e^(2x) )))) =(((1−e^(2x) )cos x+e^(2x) sinx)/((1−e^(2x) )sin x)) = cot x+(e^(2x) /(1−e^(2x) ))

$$\mathrm{y}=\mathrm{log}\sqrt{\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{1}−\mathrm{e}^{\mathrm{2x}} }}=\mathrm{log}\frac{\mathrm{sin}\:\mathrm{x}}{\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2x}} }} \\ $$$$\mathrm{y}'=\frac{\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2x}} }}{\mathrm{sin}\:\mathrm{x}}\:\:\:\frac{\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2x}} }\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2x}} }}\left(−\mathrm{2e}^{\mathrm{2x}} \right)}{\mathrm{1}−\mathrm{e}^{\mathrm{2x}} } \\ $$$$\:\:\:\:=\frac{\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2x}} }}{\mathrm{sin}\:\mathrm{x}}\:\:\frac{\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)\mathrm{cos}\:\mathrm{x}+\mathrm{e}^{\mathrm{2x}} \mathrm{sin}\:\mathrm{x}}{\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)\sqrt{\mathrm{1}−\mathrm{e}^{\mathrm{2x}} }} \\ $$$$\:\:\:\:=\frac{\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)\mathrm{cos}\:\mathrm{x}+\mathrm{e}^{\mathrm{2x}} \mathrm{sinx}}{\left(\mathrm{1}−\mathrm{e}^{\mathrm{2x}} \right)\mathrm{sin}\:\mathrm{x}} \\ $$$$\:\:\:\:=\:\mathrm{cot}\:\mathrm{x}+\frac{\mathrm{e}^{\mathrm{2x}} }{\mathrm{1}−\mathrm{e}^{\mathrm{2x}} } \\ $$

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