Задача Кузнецов Дифференцирование 12-27

Условие задачи

Найти производную.

$y={\frac {1}{\sqrt {2}}}\cdot \operatorname {arctg} {\frac {2x+1}{\sqrt {2}}}+{\frac {2x+1}{4x^{2}+4x+3}}$

Решение

$y'={\frac {1}{\sqrt {2}}}\cdot {\frac {1}{1+\left({\frac {2x+1}{\sqrt {2}}}\right)^{2}}}\cdot \left({\frac {2x+1}{\sqrt {2}}}\right)'+{\frac {(2x+1)'(4x^{2}+4x+3)-(2x+1)(4x^{2}+4x+3)'}{(4x^{2}+4x+3)^{2}}}=$

$={\frac {1}{1+{\frac {(2x+1)^{2}}{2}}}}+{\frac {2(4x^{2}+4x+3)-(2x+1)(8x+4)}{(4x^{2}+4x+3)^{2}}}={\frac {2}{4x^{2}+4x+3}}+{\frac {8x^{2}+8x+6-16x^{2}-8x-8x-4}{(4x^{2}+4x+3)^{2}}}=$

$={\frac {2(4x^{2}+4x+3)-8x^{2}-8x+2}{(4x^{2}+4x+3)^{2}}}={\frac {8}{(4x^{2}+4x+3)^{2}}}$