设f(x)=x(x-1)(x-2)…(x-1000),Z则f'(0)=
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![设f(x)=x(x-1)(x-2)…(x-1000),Z则f'(0)=](/uploads/image/z/12956790-30-0.jpg?t=%E8%AE%BEf%28x%29%3Dx%28x-1%29%28x-2%29%E2%80%A6%28x-1000%29%2CZ%E5%88%99f%27%280%29%3D)
设f(x)=x(x-1)(x-2)…(x-1000),Z则f'(0)=
设f(x)=x(x-1)(x-2)…(x-1000),Z则f'(0)=
设f(x)=x(x-1)(x-2)…(x-1000),Z则f'(0)=
f'(x)=x'(x-1)(x-2)……(x-1000)+x(x-1)'(x-2)……(x-1000)+……+x(x-1)(x-2)……(x-1000)'
除了第一项以外,其他都有x项,所以x=0时等于0
所以f'(0)=(0-1)(0-2)……(0-1000)=1000!