Question #7944d

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Answer 1 · 2017-09-21T13:24:41

$dy/dx=(-sinxcosx)/(2+cos^2x).$

$y=lnsqrt(2+cos^2x)=ln(2+cos^2x)^(1/2).$

$:. y=1/2ln(2+cos^2x)....[because, ln x^m=mlnx].$

$:. dy/dx=1/2*1/(2+cos^2x)*d/dx(2+cos^2x)..."[The Chain Rule]."$

$=1/{2(2+cos^2x)}*{d/dx2+d/dx(cosx)^2},$

$=1/{2(2+cos^2x)}{0+2cosx*d/dx(cosx)},$

$=1/{2(2+cos^2x)}{2cosx*(-sinx)},$

$rArr dy/dx=(-sinxcosx)/(2+cos^2x).$