Question

f(X)=X(X+1)(X+2)(X+3)(X+4)...X+n,则f'(0)=?

f(X)=X(X+1)(X+2)(X+3)(X+4)...X+n,则f'(0)=?

Answer 1

=n!

Answer 2

简单计算一下即可，答案如图所示

Answer 3

'(x)=x'(x+1)(x+2)…(x+n)+x(x+1)'(x+2)…(x+n)+x(x+1)(x+2)'…(x+n)+…+x(x+1)(x+2)…(x+n}'
除了第一项，后面都有因数x
则x=0时都等于0
所以f'(0)=(0+1)(0+2)…(0+n)=n！

Answer 4

f(x)=x(x+1)(x+2)(x+3)(x+4)...(x+n)
f'(x)
=(x+1)(x+2)(x+3)(x+4)...(x+n) + x(x+2)(x+3)(x+4)...(x+n)+...+
+x(x+1)(x+2)(x+3)(x+4)...(x+(n-2))(x+n)+ x(x+1)(x+2)(x+3)(x+4)...(x+(n-1))
f'(0) = (0+1)(0+2)(0+3)(0+4)...(0+n) =n!

Answer 5

f'(x)=(x+1)(x+2)(x+3)...(x+n)+x(x+2)(x+3）...(x+n)+x(x+1)(x+3)...(x+n)+...+x(x+1)(x+2)...(x+n-1)

将x=0带入上式，则
f'(0)=1*2*3*...*n=n!