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求函数y=(x-1)三次根号下x^2的极值
人气:397 ℃ 时间:2020-05-18 04:27:19
解答
y'=[(x-1)x^2/3]'
=x^2/3+(x-1)x^-1/3
=x^2/3+x^2/3-x^-1/3
=2x^2/3-x^-1/3
=2x-1/x^-1/3
J=1/2
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