求下列函数的微分：(1)y=(2x^3-3x^2+3)(根号x+1/x) (2)y=cos^3x^2
问题描述：
求下列函数的微分：(1)y=(2x^3-3x^2+3)(根号x+1/x) (2)y=cos^3x^2
答
(1)y=2x^(7/2)-2x^2-3x^(5/2)-3x+3(根号x)+3/x
dy=(7x^(5/2)-4x-(15/2)x^(3/2)-3+(3/2)(1/根号x)-3/(x^2))dx
(2)y=cos(3x^2) 吗？
dy= - 6xsin(3x^2)dx
答
(1) y = (2x^3-3x^2+3)(√x+1/x) = 2x^(7/2)-3x^(3/2)+3x^(1/2)+2x^2-3x+3/x,
dy = [7x^(5/2)-(9/2)x^(1/2)+(3/2)x^(-1/2)+4x-3-3/x^2]dx.
(2) y = (cosx^2)^3,
dy = 3[(cosx^2)^2](-sinx^2)(2x)dx.