设f(x)在上连续,在[0,π]内可导,证明至少存在一点x属于(0,π),使f'(x)=-f(x)cotx

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设f(x)在上连续,在[0,π]内可导,证明至少存在一点x属于(0,π),使f'(x)=-f(x)cotx
设f(x)在上连续,在[0,π]内可导,证明至少存在一点x属于(0,π),使f'(x)=-f(x)cotx

设f(x)在上连续,在[0,π]内可导,证明至少存在一点x属于(0,π),使f'(x)=-f(x)cotx
设g(x) = f(x)sin(x).
则g(x)在[0,π]连续, 在(0,π)可导, 且g(0) = 0 = g(π).
由Rolle定理, 存在ξ ∈ (0,π)使g'(ξ) = 0.
即有f'(ξ)sin(ξ)+f(ξ)cos(ξ) = 0.
又ξ ∈ (0,π), 故sin(ξ) ≠ 0, 有f'(ξ) = -f(ξ)cot(ξ).

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