數學挑戰題 (3) --- 微分 (易) (10分)

Question

把以下函數微分 :
(1)
y = (x - 2)(x + 1)(2x + 1)

(2)
y= 1 / 3x

(3)
y = (x + 1)(2x - 1) ^ 2

Answer 1

(1)
y = (x - 2)(x + 1)(2x + 1)
y = 2x3 - x2 - 5x - 2
dy/dx = 6x2 - 2x - 5

(2)
y = 1 / 3x
y = (1/3)(x-1)
dy/dx = (1/3)(-x-2)
dy/dx = -1/(3x2)

(3)
y = (x + 1)(2x - 1)2
y = 4x3 - 3x + 1
dy/dx = 12x2 - 3
dy/dx = 3(2x + 1)(2x - 1)

Answer 2

(1)
y=(x-2)(x+1)(2x+1)
dy/dx
=(x-2)(x+1)d(2x+1)/dx+(2x+1)d[(x-2)(x+1)]/dx
=2(x-2)(x+1)+(2x+1)[(x-2)d(x+1)/dx+(x+1)d(x-2)/dx]
=2(x-2)(x+1)+(2x+1)[(x-2)+(x+1)]
=2(x-2)(x+1)+(2x+1)(2x-1)

(2)
y=1/3x=x^(-1)/3
dy/dx=-x^(-2)/3

(3)
y=(x+1)(2x-1)^2
dy/dx=(x+1)d[(2x-1)^2]/dx+(2x-1)^2d(x+1)/dx
=(x+1)[2(2x-1)^2](2)+(2x-1)^2
=2(x+1)(2x-1)^2+(2x-1)^2
=(2x-1)^2[2(x+1)+1]
=(2x-1)^2 (2x+3)