[tex]\it (\sqrt u)' =\dfrac{u'}{2\sqrt u}\\ \\ \\ \dfrac{x}{x+1} =u\\ \\ \\ \\[/tex]
[tex]\it \Big(\sqrt{\dfrac{x}{x+1}}\Big)'=\dfrac{\Big(\dfrac{x}{x+1}\Big)'}{2\sqrt{\dfrac{x}{x+1}}}= \dfrac{\dfrac{x+1-x}{(x+1)^2}}{2\sqrt{\dfrac{x}{x+1}}}=\dfrac{1}{2(x+1)^2\sqrt{\dfrac{x}{x+1}}}[/tex]
