Question

Найти производную функции y=(2-3x)/(3x+1)

Answer 1

$y=\frac{(2-3x)}{(3x+1)};\\y'=\frac{(2-3x)'(3x+1)-(2-3x)(3x+1)'}{(3x+1)^2}=\frac{-3(3x+1)-(2-3x)3}{(3x+1)^2}=\\\frac{-9x-3-6+9x}{(3x+1)^2}=\frac{-9}{(3x+1)^2}$