а) y=x^2-x^3+ln(3x)+3/xy' = 2x - 3x^2 + 1/3x + 3*(-1/x^2) = 2x - 3x^2 + 1/3x - 3/x^2
б) y=(3x^2-1)/ln(x)y' = (3*2x)/ln(x) - (3x^2-1)/(ln(x))^2 = 6x/ln(x) - (3x^2-1)/(ln(x))^2
а) y=x^2-x^3+ln(3x)+3/x
y' = 2x - 3x^2 + 1/3x + 3*(-1/x^2) = 2x - 3x^2 + 1/3x - 3/x^2
б) y=(3x^2-1)/ln(x)
y' = (3*2x)/ln(x) - (3x^2-1)/(ln(x))^2 = 6x/ln(x) - (3x^2-1)/(ln(x))^2