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找 y' 1.e^(y*x^2) = x + y

2.找極大直
f(x) = x - e^x

3.∫[ (e^x + 1)/e^x ] dx

4.y = ln |2 - x - 5x^2 | <<<<如何變號阿~~~有絕對直ㄒ口ㄒ

2006-12-18 18:59:20 · 1 個解答 · 發問者 Xue 1 in 教育與參考 考試

就它的算法.......好了~~~怎銷絕對值

2006-12-20 04:57:28 · update #1

1 個解答

1.
e^(y*x^2) = x + y
(e^(y*x^2))(y(2x)+y'(x^2)) = 1+y'
(2xy)e^(y*x^2) + y'(x^2)e^(y*x^2) = 1+y'
y'(x^2)e^(y*x^2) - y' = 1 - (2xy)e^(y*x^2)
y' ((x^2)e^(y*x^2)-1) = 1 - (2xy)e^(y*x^2)
y' = (1-(2xy)e^(y*x^2)) / ((x^2)e^(y*x^2)-1)
2.
f(x) = x-ex
f'(x) = 1-ex
令 f'(x)=0 得 1=ex ,x=0
當 x<0 時 f'(x)>0,f(x) 為遞增
當 x>0 時 f'(x)<0,f(x) 為遞減
故知 f(x) 於 x=0 處有極大值 f(0) = 0-e0 = -1
3.
∫(ex + 1)/ex dx
=∫1+e-x dx
= x - e-x + C
4. 請問如何變號指的是?


2006-12-20 10:35:30 補充:
第4題求y'嗎?

2006-12-20 16:31:02 補充:
沒講說求什麼,就先做y'囉?
y=ln | 2-x-5x² |
因為ln的關係, 2-x-5x²不能等於0(即x不能等於(1±√(41))/(-10))
但可以大於0或者小於0
當2-x-5x²>0時
y=ln | 2-x-5x² |=ln(2-x-5x²)
y'=(1/(2-x-5x²))(-1-10x)=(-1-10x)/(2-x-5x²)
當2-x-5x²<0時
y=ln | 2-x-5x² |=ln(-(2-x-5x²))
y'=(1/(-(2-x-5x²)))(-(-1-10x))=(-1-10x)/(2-x-5x²)

2006-12-20 16:32:09 補充:
所以
y'=(-1-10x)/(2-x-5x²)

2006-12-19 16:32:48 · answer #1 · answered by chan 5 · 0 0

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