Answer
$\frac{dy}{dx}=(\frac{x-(1+x)ln(1+x)}{x^2(1+x)})(1+x)^{\frac{1}{x}}$
Work Step by Step
$y=(1+x)^{\frac{1}{x}}$
$lny=ln(1+x)^{\frac{!}{x}}$
$lny=\frac{1}{x}ln(1+x)$
$\frac{dy}{y}=(-\frac{1}{x^2}ln(1+x)+\frac{1}{x(1+x)})dx$
$\frac{dy}{dx}=(\frac{-(1+x)ln(1+x)+x}{x^2(1+x)})y$
$=(\frac{x-(1+x)ln(1+x)}{x^2(1+x)})(1+x)^{\frac{1}{x}}$$