(1) y′=2(3x+1)−(2x−3)⋅3(3x+1)2=6x+2−6x+9(3x+1)2=11(3x+1)2y' = \frac{2(3x+1)-(2x-3)\cdot 3}{(3x+1)^2} = \frac{6x+2-6x+9}{(3x+1)^2} = \frac{11}{(3x+1)^2}y′=(3x+1)22(3x+1)−(2x−3)⋅3=(3x+1)26x+2−6x+9=(3x+1)211 (2) y′=2x(3x2+1)−(x2−1)⋅6x(3x2+1)2=6x3+2x−6x3+6x(3x2+1)2=8x(3x2+1)2y' = \frac{2x(3x^2+1)-(x^2-1)\cdot 6x}{(3x^2+1)^2} = \frac{6x^3+2x-6x^3+6x}{(3x^2+1)^2} = \frac{8x}{(3x^2+1)^2}y′=(3x2+1)22x(3x2+1)−(x2−1)⋅6x=(3x2+1)26x3+2x−6x3+6x=(3x2+1)28x (3) y′=−2(2x−3)2y' = \frac{-2}{(2x-3)^2}y′=(2x−3)2−2 (4) t′=(s2+2s+2)−(s+1)(2s+2)(s2+2s+2)2=s2+2s+2−2s2−4s−2(s2+2s+2)2=−s(s+2)(s2+2s+2)2t' = \frac{(s^2+2s+2)-(s+1)(2s+2)}{(s^2+2s+2)^2} = \frac{s^2+2s+2-2s^2-4s-2}{(s^2+2s+2)^2} = \frac{-s(s+2)}{(s^2+2s+2)^2}t′=(s2+2s+2)2(s2+2s+2)−(s+1)(2s+2)=(s2+2s+2)2s2+2s+2−2s2−4s−2=(s2+2s+2)2−s(s+2)