Pr
Prombitservisnn
Ответ. y(x)=(sin(x^2-x))^2; dy(x)/dx=2*sin(x^2-x)*cos(x^2-x)*(2*x-1)=sin(2*x^2-2*x)*(2*x-1);
y=sin^2(x^2-x) ;
y' = 2sin(x^2-x)*cos(x^2-x)*(x^2-x)' ;
y' = sin(2(x^2-x))*(2x-1) ;
y' = (2x-1)*sin(2(x^2-x)).
2cos(x^2-x)*(2x-1)