Question

(x²-1)/(x+㏑x)的导函数怎么求

Answer 1

y=(x²-1)/(x+㏑x)

y'=[(x^2-1)'(x+lnx)-(x^2-1)(x+lnx)']/(x+lnx)^2

=[2x(x+lnx)-(x^2-1)(1+1/x)]/(x+lnx)^2

=(x^2+2xlnx-x-1/x+1)/(x+lnx)^2