如图一题~求函数的n阶导数
1个回答
2016-11-06
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y=(x-1)/(x+1) = {(x+1)-2}/(x+1) = 1 - 2/(x+1)
y ′ = 2/(x+1)²
y ′′ = -4/(x+1)³
......
y(n) = (-1)^(n+1) * 2 * n! /(x+1)^(n+1)
y ′ = 2/(x+1)²
y ′′ = -4/(x+1)³
......
y(n) = (-1)^(n+1) * 2 * n! /(x+1)^(n+1)
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