Question

найти производную: y=arcctg*1+z/1-z

Answer 1

$(arctanfrac{1+z}{1-z})'_z=frac{1}{1+(frac{1+z}{1-z})^2}*frac{1*(1-z)-(-1)(1+z)}{(1-z)^2}=$

$=frac{1}{1+(frac{1+z}{1-z})^2}*frac{1-z+1+z}{(1-z)^2}=frac{2}{(1-z)^2+(1+z)^2}=$

$=frac{2}{2+2z^2}=frac{1}{1+z^2}$

Ответ: $frac{1}{1+z^2}$