y=(1-sinx)/(1+sinx) производную

Формулы:

$(\frac{u}{v})' = \frac{u'v+uv'}{v^{2}}\\\\ (u+/-v)'=u'+/-v'$

$y=\frac{1-sinx}{1+sinx}\\ y'=(\frac{1-sinx}{1+sinx})'=\frac{(1-sinx)'\cdot(1+sinx)-(1-sinx)\cdot(1+sinx)'}{(1+sinx)^{2}}=\\\\ =\frac{-cosx\cdot(1+sinx)-(1-sinx)\cdot cosx}{(1+sinx)^{2}}=\frac{-cosx-cosx\cdot sinx-cosx+cosx\cdot sinx}{(1+sinx)^{2}}=\\\\ =-\frac{2cosx}{(1+sinx)^{2}}$