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发表于 5-3-2006 06:55 PM
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怎么你问的问题跟我的一个朋友问的一摸一样?哈哈
我们就来证明 d(ln x)/dx = 1/x ( 酱的话,当你 integrate both side 就会得到 kamir 1/x = ln x)
letting f(x) = ln x , by first principle
f'(x) = lim_{h->0} (f(x+h)-f(x))/h = lim_{h->0} (ln(x+h)-ln x)/h
= lim_{h->0} 1/h * ln ( 1 + h/x )
= lim_{h->0} ln (1 + h/x)^(1/h)
but by definition of e , we have
lim_{h->0} ( 1 + h)^(1/h) = e
因此 lim_{h->0} ln (1 + h/x)^(1/h) = lim_{h->0} ln (1 + h/x)^{(1/(h/x))1/x} = ln e^(1/x) = 1/x
所以 d(ln x)/dx = f'(x) = 1/x
$ 1/x dx = ln x + C ( $ = integrate ) |
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