✔ 最佳答案
Q1:
f(x)=∫[0~x] √[1+(cost)^2] dt
f'(x)= √[1+(cosx)^2 ]
設g(x)為 f(x)之反函數f^(-1)(x), 則
g'(f(x))= 1/ f'(x)
(1)g'(A)= 1/f'(π/2)= 1/√2
(2)g'(B)= 1/f'(5π/6)=1/√[1+(-√3 /2)^2] = 2/√7
(3)f(0)=0, g'(0)=1/f'(0)= 1/√2
Q2:
d/dx (x^x)= d/dx[ exp(x lnx)]= exp(x lnx)*(1+lnx)= x^x (1+lnx)
d/dx (x^x)^x = d/dx [ x^(2x)]= x^(2x) * (2+ 2lnx)
Note:題目應是 d/dx x^(x^x)吧!?
d/dx [x^(x^x)] = d/dx [ exp( x^x * lnx) ]
= exp(x^x * lnx) *[ x^x (1+lnx) * lnx + x^(x-1)]
= x^(x^x) * x^x ( 1+ lnx + 1/x)
Q3:
(a) lim(x->0+) f(x)= lim(y-> -∞) y/(1+y^2) = 0
Note: lim(x->0+) ln(x)= - ∞
(b) f(x) = y/(1+y^2) 設為 g(y), y= lnx= -∞ ~ ∞
g'(y)= (1-y^2)/(1+y^2)^2= (1+y)(1-y)/(1+y^2)^2
=> -1 < y < 1時 g'(y) >0 , g(y)遞增
|y| > 1時, g(y)遞減
又lim(y->∞) g(y)= 0 = lim(y->-∞) g(y)
故 f(x)最大值= g(1)= 1/2
f(x)最小值= g(-1)= -1/2
(c)F(x)=∫[1~ x^2] f(t) dt
F'(x)= f(x^2) * 2x (Fundamental Theorem of Calculus)
F'(√e) = f(e)* 2√e = (1/2) * 2√e = √e
2009-07-09 12:30:36 補充：
d/dx (x^x)^x = d/dx [ x^(x^2)]= d/dx exp(x^2 lnx)
= exp(x^2 lnx)*(2x lnx + x)= x (2lnx +1) (x^x)^x
Note: exp(x)= e^x
