设f(x)=x(x-1)(x-2)……(x-100)求,f'(1)
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设f(x)=x(x-1)(x-2)……(x-100)求,f''(1)设f(x)=x(x-1)(x-2)……(x-100)求,f''(1)设f(x)=x(x-1)(x-2)……(x-100)求,f''(1)f
设f(x)=x(x-1)(x-2)……(x-100)求,f'(1)
设f(x)=x(x-1)(x-2)……(x-100)求,f'(1)
设f(x)=x(x-1)(x-2)……(x-100)求,f'(1)
f(x)=x(x-1)(x-2)……(x-100)
构造函数,设g(x)=x-1,h(x)=x(x-2)……(x-100)
所以,f(x)=g(x)·h(x)
f'(x)
=g'(x)h(x)+g(x)h'(x)
x=1时,g(1)=0,g(x)h'(x)=0,不需要对h'(x)进行求解
所以,
f'(1)
=g'(1)h(1)
=1×[1×(1-2)……×(1-100)]
=（-1）×99!
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