## Calculus: Early Transcendentals (2nd Edition)

$f'(x)=\frac{2}{(x+2)^2}$ $f''(x)=\frac{-4}{(x+2)^3}$
$f(x)=\frac{x}{x+2}$ Using Quotient Rule, where $(\frac{f}{g})'=\frac{f'g-fg'}{g^2}$ $f'(x)=\frac{(1)(x+2)-(x)(1)}{(x+2)^2}=\frac{2}{(x+2)^2}$ Using Quotient Rule and Chain Rule: $f''(x)=\frac{(0)(x+2)^2-2(2)(1)(x+2)}{(x+2)^4}=\frac{-4}{(x+2)^3}$