一道数列问题 设f(x)=x(x-1)(x-2)...(x-100),则f‘(x)等于? 10
2011-08-07
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呃,是一阶导数啊
把后面的(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)]‘
同样的把(x-2)...(x-100)再次看成一个整体求导
一步一步的类推
f'(x)=(1/x+1/(x-1)+1/(x-2)+......+1/(x-100))x[(x-1)(x-2)...(x-100)]
用隐函数求导的话可以用
对数(很常用的)
y=f(x)=x(x-1)(x-2)……(x-100)
lny=lnx+ln(x-1)+ln(x-2)+……+1n(x-99)+ln(x-100)
y'/y=(1/x+1/(x-1)+1/(x-2)+......+1/(x-100)
y'=(1/x+1/(x-1)+1/(x-2)+......+1/(x-100))x[(x-1)(x-2)...(x-100)]
把后面的(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)]‘
同样的把(x-2)...(x-100)再次看成一个整体求导
一步一步的类推
f'(x)=(1/x+1/(x-1)+1/(x-2)+......+1/(x-100))x[(x-1)(x-2)...(x-100)]
用隐函数求导的话可以用
对数(很常用的)
y=f(x)=x(x-1)(x-2)……(x-100)
lny=lnx+ln(x-1)+ln(x-2)+……+1n(x-99)+ln(x-100)
y'/y=(1/x+1/(x-1)+1/(x-2)+......+1/(x-100)
y'=(1/x+1/(x-1)+1/(x-2)+......+1/(x-100))x[(x-1)(x-2)...(x-100)]
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等于每一项的导数乘以剩余的项 的和
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