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求y=ln(x+根号下x^2+1)函数的导数
人气:158 ℃ 时间:2019-07-29 19:34:28
解答
y'=1/(x+根号下x^2+1)*(x+根号下x^2+1)'
=1/(x+根号下x^2+1)*(1+x/根号下x^2+1)
=1/(x+根号下x^2+1)*(根号下x^2+1+x)/根号下x^2+1
=1/根号下(x^2+1)
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