## Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{2-x}{2\sqrt{x}(2+x)^{2}}$
$y=\dfrac{\sqrt{x}}{2+x}$ Differentiate using the quotient rule: $y'=\dfrac{(2+x)(\sqrt{x})'-(\sqrt{x})(2+x)'}{(2+x)^{2}}=...$ Rewrite the square root using fractionary power and continue with the differentiation process: $...=\dfrac{(2+x)(x^{1/2})'-(x^{1/2})(2+x)'}{(2+x)^{2}}=...$ $=\dfrac{(2+x)(\dfrac{1}{2}x^{-1/2})-(x^{1/2})(1)}{(2+x)^{2}}=...$ Evaluate the products and simplify: $...=\dfrac{x^{-1/2}+\dfrac{1}{2}x^{1/2}-x^{1/2}}{(2+x)^{2}}=\dfrac{x^{-1/2}-\dfrac{1}{2}x^{1/2}}{(2+x)^{2}}=\dfrac{\dfrac{1}{\sqrt{x}}-\dfrac{\sqrt{x}}{2}}{(2+x)^{2}}=\dfrac{\dfrac{2-x}{2\sqrt{x}}}{(2+x)^{2}}=\dfrac{2-x}{2\sqrt{x}(2+x)^{2}}$