大學微積分的問題

f(x)=x√(x^2-1)
f’(x) = √(x^2-1) + x/√(x^2-1) = (x^2+x-1)/√(x^2-1)
f”(x)
= {(2x +1)*√(x^2-1) – (x^2+x-1)* [x/√(x^2-1)]}/(x^2-1)
=(x^3-x-1)/(x^2-1)
當f”(x) =0時,x≒1.325

f(x)=x/(x^2+3)
f’(x) = [(x^2+3) – 2x^2]/(x^2 +3)^2 = (-x^2 +3)/(x^2+3)^2
f”(x)
=[-2x(x^2+3) – 4x(x^2+3)(-x^2+3)]/(x^2+3)^4
=[4x^3 –14x]/(x^2+3)^3
當f”(x) =0時,x = 0,√14/2, -√14/2

y+xy=4
dy +ydx + xdy = 0
(1+x)dy = -ydx
dy/dx = -(1+x)/y
當(x,y) = (-5,-1)時,dy/dx = -4………..(解答)


2009-10-23 18:16:01 補充：
嗯…….果然有作錯的地方,重新再來!!
f(x)=x√(x^2-1)
f’(x) = √(x^2-1) + x^2/√(x^2-1) = (2x^2 –1)/ √(x^2-1)
f”(x)
= [4x*√(x^2-1) – (x(2x^2-1)/√(x^2-1))]/(x^2-1)
= (2x^3 –3x)/(x^2-1)
當f”(x) = 0時,x = 0,√6/2, -√6/2

2009-10-23 18:16:06 補充：
f(x)=x/(x^2+3)
f’(x) = [(x^2+3) – 2x^2]/(x^2+3)^2 = (-x^2+3)/(x^2+3)^2
f”(x)
= [-2x*(x^2+3)^2 -4x(x^2+3)(-x^2+3)]/(x^2+3)^4
=[ -2x(x^2+3) –4x(-x^2+3)]/(x^2+3)^3
=(2x^3 –18x)/(x^2+3)^3
當f”(x) =0時,x = 0,3,-3
感謝nelsonywm2000大大無私的提醒!!

2009-10-24 19:04:23 補充：
f”(x) = 0時,2x^3-3x = 0
x(2x^2 –3) = 0
2x^2 –3 =0,則x^2 = 3/2,x = √6/2或 -√6/2
你大概是換算時,把分母的√2看成是2了吧!!

2009-10-25 18:41:58 補充：
3/2^1/2的意思是上下都要開根號!!
所以3/2^1/2 = √3/√2 = √6/2
這樣應該沒問題吧!!這是國中最基本的東西喔!!

請對一下,39及41都做錯了