Найти производную: y=arcctg*1+z/1-z

[latex](\arctan\frac{1+z}{1-z})'_z=\frac{1}{1+(\frac{1+z}{1-z})^2}*\frac{1*(1-z)-(-1)(1+z)}{(1-z)^2}=[/latex] [latex]=\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}=[/latex] [latex]=\frac{2}{2+2z^2}=\frac{1}{1+z^2}[/latex] Ответ: [latex]\frac{1}{1+z^2}[/latex]