求y=ln(x^2+根号下(x^2+1))的微分_数学解答

求y=ln(x^2+根号下(x^2+1))的微分

人气：207 ℃　时间：2020-01-04 01:34:40

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)