petua hasil darab [product rule]
Jika [If] y=uv
(dy/dx) = u(dv/dx) + v(du/dx)
or
f ' (x) = u(dv/dx) + v(du/dx)
f(x) = (2x3-3)[(1/x)-2x]2
.
Let u = (2x3-3) , v = [(1/x)-2x]2
(du/dx) = (6x2) , (dv/dx) = (8x4-2)/x3
.
(note: anxn-1)
2x3-3 = 2(3)x3-1 - 3(0)
= 6x2
.
[(1/x)-2x]2= (2)[(1/x)-2x]2-1 [-(1/x2) - 2]
= (2) [(1/x)-2x] [-(1/x2)-2]
= (2) [(1-2x2)/x] [-(1+2x2)/x2]
= -[(2)(1-2x2)(1+2x2)] / [(x)(x2)]
= -[2(12-[2x2]2) / x3]
= -[2(1-4x4) / x3]
= -[2-8x4 / x3]
= [8x4-2 / x3]
f ' (x) = u(dv/dx) + v(du/dx)
= (2x3-3) [(8x4-2)/x3] + [(1/x)-2x]2 (6x2)
= [(2x3-3)(8x4-2)] / x3 + (6x2) [(1-2x2)/x]2
= (16x7-24x4-4x3+6) / x3 + (6x2) [(1-2x2)2/x2)]
= (16x7-24x4-4x3+6) / x3 + (6)(-2x2+1)2
= (16x7-24x4-4x3+6) / x3 + (6)(4x4-4x2+1)
= (16x7-24x4-4x3+6) / x3 + (24x4-24x2+6)
= (16x7-24x4-4x3+6) + x3(24x4-24x2+6) / x3
= (16x7-24x4-4x3+6)+(24x7-24x5+6x3) / x3
= (40x7 -24x4 +2x3 +6 -24x5) / x3
= 40x4 -24x +2 +(6/x3) -24x2 [重新排列]
= 40x4 - 24x2 -24x + (6/x3) +2 [答案]
(2x3-3)(8x4-2)
= (2x3)(8x4) + (-3)(8x4) + (2x3)(-2) + (-3)(-2)
= 16x7 -24x4 -4x3 +6
1和-2x2交换位子
a2-2ab+b2 = (-2x2+1)2
= (-2x2)2 +1(-2x2)(2) +12)
= 4x4 -4x2 + 1
变分数乘x3
上面分子的x都减x3(分母)
example : x4/x3 = x4-3 =x
