y=ln[arctan1/(1+x)]的导数
人气:133 ℃ 时间:2020-04-22 20:14:21
解答
y=ln[arctan(1/(1+x))]y' = (1/[arctan(1/(1+x))])[arctan(1/(1+x))]'=(1/[arctan(1/(1+x))]).[ (1+x)^2/(1+(1+x)^2)] .(1/(1+x))'=(1/[arctan(1/(1+x))]).[ (1+x)^2/(1+(1+x)^2)] .(-1/(1+x)^2)=(1/[arctan(1/(1+...
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