设函数f(x)=x*(x-1)*(x-2)...(x-99)(x-100),则f′(0)=
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设函数f(x)=x*(x-1)*(x-2)...(x-99)(x-100),则f′(0)=
因为f(x)=x*(x-1)*(x-2)...(x-99)(x-100)
所以f(0)=0
所以f′(0)=lim(x→0)[f(x)-f(0)]/(x-0)
=lim(x→0)f(x)/x
=lim(x→0)[x*(x-1)*(x-2)...(x-99)(x-100)]/x
=lim(x→0)(x-1)*(x-2)...(x-99)(x-100)
=(0-1)*(0-2)...(0-99)(0-100)
=100!
注n!=1*2*3*...*n(!是阶乘的意思)
所以f(0)=0
所以f′(0)=lim(x→0)[f(x)-f(0)]/(x-0)
=lim(x→0)f(x)/x
=lim(x→0)[x*(x-1)*(x-2)...(x-99)(x-100)]/x
=lim(x→0)(x-1)*(x-2)...(x-99)(x-100)
=(0-1)*(0-2)...(0-99)(0-100)
=100!
注n!=1*2*3*...*n(!是阶乘的意思)
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