一題重積分

f(x) = ∫[0,x^2] ∫[0,v^2] -exp(-u) du dv
=∫[0, x^2] [exp(-v^2) -1 ]dv
f'(x) = [exp(-x^4) - 1]* 2x ( Fundamental theroem of Calculus)
=> f'(1) = 2 ( 1/e - 1)

2009-02-15 19:46:57 補充：
微積分基本定理:
F(x) = ∫[u(x), v(x)] f(t) dt
=> F'(x) = f(v(x))*v'(x) - f(u(x))*u'(x)

Fundamental theroem of Calculus
呵...大師你的文言文難倒人囉
微積分基本定理就是找到其(積分式)的反導數再代上下限
稱為微積分基本定理(Fundamental theroem of Calculus)