原帖內容
mejack1324

2009-11-27 14:31


1a)   Let y = (x^3) / (x^2) + 1  <---可約簡= = '?
          then y = x+1
Differentiate  with respect to x
  dy/dx = 1

1b)  y= (1/2)sinx
Differentiate  with respect to x
dy/dx = 1/2 *[ d (sinx)/ dx]
dy/dx = 1/2 cosx

When  x= (1/3)兀 (您可看做 183/3=60)
dy/dx = 1/2 cos((1/3)兀) = 1/2 * 1/2 = 1/4


the rate of increase is zero . then put dy/dx = 0
dy/dx = 1/2 cosx = 0
1/2 cosx = 0
cosx = 0

for 0<=x <= 2兀 <---( 題目應該有 range 的. 否則幾細都可以)
cosx = 0
x=兀/2

1c)  y= 4+7x-(3x^3)
        dy/dx = 0 + 7 - 9x^2 = 7-9x^2
         d^2y / dx ^2= 0 - 18x = -18x

dy/dx = 7-9x^2
d^2y / dx ^2 = -18x

when (d^2)y/dx^2 =9
-18x=9
x= -1/2

put x = -1/2 into dy/dx = 7-9x^2
dy/dx = 7 - 9(-1/2)^2
= 7 - 9/4
=19/4

希望沒有錯 0.0

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威望 +2 生神仙 2009-12-2 12:00 熱心幫助人