Question

麻烦帮忙求个导数看

图示函数，麻烦给个求导解答

Answer 1

y=[1+(lnx)^2] ^(1/3)

y'
=(1/3)[1+(lnx)^2]^(-2/3) .[1+(lnx)^2]'
=(1/3)[1+(lnx)^2]^(-2/3) .(2lnx) . (lnx)'
=(1/3)[1+(lnx)^2]^(-2/3) .(2lnx) . (1/x)
=(2/3) lnx/ {x.[1+(lnx)^2]^(2/3) }