Answer
$y'=\dfrac{1}{2}\dfrac{1}{(x+1)^{2}}\sqrt{\dfrac{x+1}{x}}$
Work Step by Step
$y=\sqrt{\dfrac{x}{x+1}}$
Let's rewrite the function like this:
$y=(\dfrac{x}{x+1})^{1/2}$
Differentiate using the chain rule:
$y'=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}(\dfrac{x}{x+1})'=...$
Use the quotient rule to find $(\dfrac{x}{x+1})'$:
$...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}[\dfrac{(x+1)(x)'-(x)(x+1)'}{(x+1)^{2}}]=...$
$...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}[\dfrac{(x+1)(1)-(x)(1)}{(x+1)^{2}}]=...$
$...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}\dfrac{x-x+1}{(x+1)^{2}}=...$
$...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}\dfrac{1}{(x+1)^{2}}=\dfrac{1}{2}\dfrac{1}{(x+1)^{2}}\sqrt{\dfrac{x+1}{x}}$
