## Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{1}{2}\dfrac{1}{(x+1)^{2}}\sqrt{\dfrac{x+1}{x}}$
$y=\sqrt{\dfrac{x}{x+1}}$ 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 $(\dfrac{x}{x+1})'$: $...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}[\dfrac{(x+1)(x)'-(x)(x+1)'}{(x+1)^{2}}]=...$ $...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}[\dfrac{(x+1)(1)-(x)(1)}{(x+1)^{2}}]=...$ $...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}\dfrac{x-x+1}{(x+1)^{2}}=...$ $...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}\dfrac{1}{(x+1)^{2}}=\dfrac{1}{2}\dfrac{1}{(x+1)^{2}}\sqrt{\dfrac{x+1}{x}}$