Question

50 баллов!!! Найти производную

Answer 1

$y' = \frac{-sinx*sin^2x-2sinxcosx*cosx}{sin^4x} +2sinx- \frac{3}{2cos^2 \frac{x}{2} tg \frac{x}{2} } =\\\\= \frac{-1-cos^2x}{sin^3x} +2sinx- \frac{3}{2cos \frac{x}{2}sin \frac{x}{2} } = \frac{-1-cos^2x+2sin^4x-3sin^2x}{sin^3x} =\\\\= \frac{2sin^4x-2sin^2x-2}{sin^3x}$