作者LuisSantos (^______^)
看板trans_math
标题Re: [积分]
时间Fri Apr 27 11:27:39 2007
※ 引述《king911015 (早已放弃爱上你)》之铭言：
: 1.
: Let g be differentiable real-valued function satisfied
: 2x 2x
: (∫ g(t) dt ) sinx + cosx^2 = x + (∫g(t) dt ) for all x near 0. Find g'(0) =?
: 0 0
2x 2x
(∫ g(t) dt)(sinx) + (cosx)^2 = x + (∫ g(t) dt)
0 0
等号两边对x微分
2x
(g(2x))(2)(sinx) + (∫ g(t) dt)(cosx) + (2)(cosx)(-sinx) = 1 + (g(2x))(2)
0
2x
(2)(g(2x))(sinx) + (∫ g(t) dt)(cosx) - sin2x = 1 + (2)(g(2x)) ------(1)
0
2x = 0 => x = 0
x = 0 代入上式得
(2)(g(0))(0) + (0)(1) - 0 = 1 + (2)(g(0))
(2)(g(0)) + 1 = 0
-1
(2)(g(0)) = -1 => g(0) = ---
2
(1)的等号两边对x微分
2x
(4)(g'(2x))(sinx) + (2)(g(2x))(cosx) + (g(2x))(2)(cosx) + (∫ g(t) dt)(-sinx)
0
- (2)(cos2x) = (4)(g'(2x))
2x
(4)(g'(2x))(sinx) + (4)(g(2x))(cosx) - (∫ g(t) dt)(sinx) - (2)(cosx2x)
0
= (4)(g'(2x)) ------(2)
2x = 0 => x = 0 代入(2)得
(4)(g'(0))(0) + (4)(g(0))(1) - 0 - 2 = (4)(g'(0))
-1
(4)(---)(1) - 2 = (4)(g'(0))
2
(4)(g'(0)) = -4 => g'(0) = -1
--
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