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求y=[ln(1-x)^2]^2的微分
人气:187 ℃ 时间:2020-01-30 00:33:05
解答
y=[ln(1-x)^2]^2
y'= 2[ln(1-x)^2] * [ln(1-x)^2]'
= 2[ln(1-x)^2] * [2ln(1-x)]'
= 2[ln(1-x)^2] * 2*1/(1-x)
= 4*[ln(1-x)^2]/(1-x)
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