设f(x)=xe-x,求fn(x).

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问题 设f(x)=xe-x,求fn(x).

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答案fˊ(x)=e-x+e-x(-1)x =(1-x)e-x =-(x-1)e-x f〞(x)=-e-x+xe-x-e-x =(-1)2(x-2)e-x f’’’(x)=(-1)3(x-3)e-x f(k)(x)=(-1)k(x-k)e-x f(k+1)(x)=((-1)k(x-k)e-x)ˊ =(-1)k[e-x+(x-k)(-e-x)] =(-1)k+1(x-(k+1))e-x 由数学归纳法知 f(n)(x)=(-1)n(x-n)e-x

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