设函数f(x)=(x-1)(x-2)...(x-100)(x>100),求F'(X)

设函数f(x)=(x-1)(x-2)...(x-100)(x>100),求F'(X)
设函数f(x)=(x-1)(x-2)...(x-100)(x>100),求F'(X)

设函数f(x)=(x-1)(x-2)...(x-100)(x>100),求F'(X)
lnf(x)=ln(x-1)+ln(x-2).+ln(x-100)
两侧求导得：f'(x)/f(x)=1/(x-1)+1/(x-2)+...+1/(x-100)
f'(x)=(x-1)(x-2)...(x-100)[1/(x-1)+1/(x-2)+...+1/(x-100)]

f'(X)=(x-1)(x-2)...(x-100)[1/(x-1)+1/(x-2)+...+1/(x-100)]=f(x)[1/(x-1)+1/(x-2)+...+1/(x-100)]

他应该不会要你求f（x）的导数的表达式，应该是求一个具体的数，也就是说最后一个括号里的x是一个常数。如果是这样，不要用公式，尝试一下求导的定义式，应该会得到答案。