若f(X)=X(X-1)(X-2).(X-100),则f‘(0)=?

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若f(X)=X(X-1)(X-2).(X-100),则f‘(0)=?
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若f(X)=X(X-1)(X-2).(X-100),则f‘(0)=?
若f(X)=X(X-1)(X-2).(X-100),则f‘(0)=?

若f(X)=X(X-1)(X-2).(X-100),则f‘(0)=?
把(x-1)(x-2)...(x-100)看成一个整体
f(x)=x(x-1)(x-2)...(x-100)
f'(x)=x ' *[(x-1)(x-2)...(x-100)]+x[(x-1)(x-2)...(x-100)]‘
=(x-1)(x-2)...(x-100)+x[(x-1)(x-2)...(x-100)]‘
f‘(0)=100!

f‘(x)=(X-1)(X-2)...........(X-100),则f‘(0)=100!

f(x)=x(x-1)(x-2)...(x-100)
f'(x)=x ' *[(x-1)(x-2)...(x-100)]+x[(x-1)(x-2)...(x-100)]‘
=(x-1)(x-2)...(x-100)+x[(x-1)(x-2)...(x-100)]‘
f‘(0)=100

f'(x)=(x-1)(x-2)……(x-100)+x(x-2)(x-3)……(x-100)+……+x(x-1)(x-2)……(x-99)
故f'(0)=100!
也就是1一直乘到100,因为f‘(x)后面几项都乘以x,也就是f'(0)时这几项都是0

令 g(x) = (x-1)(x-2)...........(x-100)

则 f(x) = xg(x)

由两函数相乘求导公式可知,f'(x) = g(x) + xg'(x)

x=0时,xg'(x)一项为0

所以 ...

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令 g(x) = (x-1)(x-2)...........(x-100)

则 f(x) = xg(x)

由两函数相乘求导公式可知,f'(x) = g(x) + xg'(x)

x=0时,xg'(x)一项为0

所以 f'(0) = g(0) = (-1)(-2)...(-100) = 100!

(! 表示 阶乘)


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做法好像是你先拆开前面一两个,然后推出规律

如果先把0带入,在求导的话答案不是0么