Question

derivative of( cos2x+1)/sin2x​

Answer 1

$\frac{-2(1+cos2x)}{sin^{2}2x }$

Step-by-step explanation:

$\frac{u}{v} \\=\frac{u'.v-v'u}{u^{2} } \\=\frac{(-2.sin2x).sin2x-2cos2x(cos2x+1)}{(sin^{2}2x) } \\=\frac{-2(sin^{2}2x)-2cos^{2}2x-2cos2x }{sin^{2}2x} \\=\frac{-2(sin^{2}2x+cos^{2}2x)-2cos2x}{sin^{2}2x}\\=\frac{-2-2cos2x}{sin^{2}2x}\\=\frac{-2(1+cos2x)}{sin^{2}2x}$