[tex]\it f(x) = 2ln(x+1)\\ \\ f'(x) = \dfrac{2}{x+1}[/tex]
[tex]\Big[2\ln(x+1)\Big]' = 2\cdot \dfrac{1}{x+1}\cdot (x+1)' = \dfrac{2}{x+1}\cdot 1 = \dfrac{2}{x+1}[/tex]
[tex]\boxed{(\ln u)' = \dfrac{1}{u}\cdot u'}\text{ - formula}[/tex]
