Question

Найти производную функции y = 1-sinx/1+cosx

Answer 1

$y = \frac{1-sin(x)}{1+cos(x)}\\\\ y' = \frac{(1-sin(x))'\cdot (1+cos(x)) - (1+cos(x))'\cdot (1-sin(x))}{(1+cos(x))^2}= \\\\ = \frac{(-cos(x))\cdot (1+cos(x)) - (-sin(x))\cdot (1-sin(x))}{(1+cos(x))^2} = \\\\ = \frac{-cos(x)-cos^2(x)+sin(x)-sin^2(x)}{(1+cos(x))^2}= \\\\ = \frac{sin(x)-cos(x)-1}{(1+cos(x))^2}=-\frac{1+cos(x)-sin(x)}{(1+cos(x))^2} =\\\\ = -\frac{1}{1+cos(x)} + \frac{sin(x)}{(1+cos(x))^2}.$