Question

F(x) = (lnx) / (x + 1) f ' (e)

Answer 1

$f(x)=\frac{lnx}{x+1}\\\\f`(x)=\frac{\frac{1}{x}(x+1)-lnx}{(x+1)^2}=\frac{x+1-x\cdot lnx}{x(x+1)^2}=\frac{x(1-lnx)+1}{x(x+1)^2}\\f`(e)=\frac{e(1-lne)+1}{e(e+1)^2}=\frac{1}{e(x+1)^2}\; \; \; \; [lne=1\; \to \; (1-lne)=0]$