Answer
$\frac{d}{dx}\left(\frac{\ln(x)}{\ln(x)+1}\right) = \frac{1}{x(\ln(x)+1)^2}$
Work Step by Step
Using Chain Rule and Quotient Rule
$\frac{d}{dx}(\frac{ln(x)}{ln(x)+1}) = \frac{(\frac{1}{x})(ln(x)+1)-(ln(x))(\frac{1}{x})}{(ln(x)+1)^2} = \frac{1}{x(\ln(x)+1)^2}$