m2 differential

用戶31980

2011-05-02 1:12 am

find dy/dx,
1.y=(x+1)((x+2)^2)((x+3)^2))
ans:(x^2+5x+6)^2+2(x+1)(2x+5)(x^2+5x+6)
2.y=1/(x+(x^2-1)^1/2)
ans:1-(x/(x^2-1)^1/2)

更新1:

please show the steps

更新2:

myisland8132 sorry, i do not quite understand q 2 can you use the quotient rule to solve it?

更新3:

please use a clearer method to present the answer or use quotient rule, otherwise i am going 移除...................

回答 (2)

rex

2011-05-06 4:46 am

✔ 最佳答案

1.連續乘積我們可以利用對數來作
y=(x+1)(x+2)^2(x+3)^2
lny=ln(x+1)+2ln(x+2)+2ln(x+3)兩端微分
(1/y)dy/dx=1/(x+1)+2/(x+2)+2/(x+3)
dy/dx=y[1/(x+1)+2/(x+2)+2/(x+3)]
=(x+2)^2(x+3)^2+2(x+1)(x+2)(x+3)^2+2(x+1)(x+2)^2+(x+3)

2.y=1/(x+√(x^2-1)) --> d(x+√(x^2-1))/dx=1+x/√(x^2-1)=(x+√(x^2-1))/√(x^2-1)
dy/dx=-(x+√(x^2-1))/[√(x^2-1)(x+√(x^2-1))^2]
=-1/[√(x^2-1)(x+√(x^2-1)], since 1=x^2-[√(x^2-1)]^2
=-[x-√(x^2-1)]/√(x^2-1)
=1-x/√(x^2-1)

2011-05-05 21:26:59 補充：
1.最後一行一點小筆誤
=(x+2)^2(x+3)^2+2(x+1)(x+2)(x+3)^2+2(x+1)(x+2)^2(x+3)

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myisland8132

2011-05-02 2:22 am

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

dy/dx

= 2(x + 1)(x + 2)^2(x + 3) + 2(x + 1)(x + 3)^2(x + 2) + [(x + 2)(x + 3)]^2

= 2(x +1)(x + 2)(x + 3)(x + 2 + x + 3) + [(x + 2)(x + 3)]^2

= (x^2 + 5x + 6)^2 + 2(x + 1)(2x + 5)(x^2 + 5x + 6)

2 y = 1/[x + √ (x^2 - 1)]

dy/dx

= -1/[x + √ (x^2 - 1)]^2 * [1 + x/√ (x^2 - 1)]

= -1/[x + √ (x^2 - 1)]^2 * [(x + √ (x^2 - 1) ) /√ (x^2 - 1)]

= -1/[x^2 - 1 + x√ (x^2 - 1)]

= -[x^2 - 1 - x√ (x^2 - 1)]/[x^2 - 1 + x√ (x^2 - 1)]

= -[x^2 - 1 - x√ (x^2 - 1)]/[x^4 - 2x^2 + 1 - x^4 + x^2]

= -[x^2 - 1 - x√ (x^2 - 1)]/[1 - x^2]

= 1 - x/√ (x^2 - 1)

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收錄日期: 2021-04-26 19:17:26
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