Differentiation

2007-12-20 12:38 am
Differentiate the function with respect to x,

(2x-1)^3(4-x)^4

用chain rule 計呀~我唔識~

回答 (3)

2007-12-20 1:23 am
✔ 最佳答案
先抄問題,
(2x-1)^3(4-x)^4
後做低佢,chain rule記住:
前面唔攪, d後面;加,後面唔攪,d前面
前面唔攪, d後面:
(2x-1)^3(4)(4-x)^3(-1)

後面唔攪,d前面
(4-x)^4(3)(2x-1)^2(2)
即係:
(2x-1)^3(4)(4-x)^3(-1)+(4-x)^4(3)(2x-1)^2(2)
=-4(4-x)^3(2x-1)^3+6(4-x)^4(2x-1)^2
攪掂

2007-12-19 17:39:11 補充:
想令仔d就秋埋答案:-4(4-x)^3(2x-1)^3+6(4-x)^4(2x-1)^2=2(4-x)^3(2x-1)^2〔-2(2x-1)+3(4-x)〕=2(4-x)^3(2x-1)^2〔-4x+2+12-3x〕=2(4-x)^3(2x-1)^2(14-7x)=14(4-x)^3(2x-1)^2〔2-x)令仔哂~

2007-12-19 17:42:10 補充:
答案最令係我
2007-12-20 9:52 pm
Let y=(2x-1)^3(4-x)^4
dy/dx=(2x-1)^3[4(4-x)^3](-1)+(4-x)^4[3(2x-1)^2](2)
=-4(2x-1)^3(4-x)^3+6(4-x)^4(2x-1)^2
=2(2x-1)^2(4-x)^3[3(4-x)-2(2x-1)]
=2(2x-1)^2(4-x)^3[(14-7x)]
=14(2x-1)^2(4-x)^3(2-x)
2007-12-20 1:12 am
d/dx[(2x-1)^3(4-x)^4]
=[(2x-1)^3]d/dx[(4-x)^4]+[(4-x)^4]d/dx[(2x-1)^3]
=[(2x-1)^3][4(4-x)^3]d/dx(4-x)+[(4-x)^4][3(2x-1)^2]d/dx(2x-1)
=-[(2x-1)^3][4(4-x)^3]+2[(4-x)^4][3(2x-1)^2]
=(2x-1)^2(4-x)^3[-4(2x-1)+6(4-x)]
=(2x-1)^2(4-x)^3(-14x+28)
=-14(2x-1)^2(4-x)^3(x-2)

2007-12-19 17:14:05 補充:
上面個位做錯數


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