Use logarithmic differentiation to find dy/dx for y=(1+x)^(1/x).

Answer 1

ln y = ln (1+x)^(1/x)
= (1/x) ln (1+x)
(dy/dx) /y = (-1/x^2)(ln(1+x) + (1/x)(1/(1+x)

dy/dx = y [(-1/x^2) ( ln(1+x) + x/(1+x) ]
or (-1/x^2)( ln(1+x) + x/(1+x) ) (1+x)^(1/x)

pretty messy looking ...