## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{1}{(a+1)(a+h+1)}$
$f(x)=\dfrac{x}{x+1}$ Find $f(a)$ by substituting $x$ with $a$: $f(a)=\dfrac{a}{a+1}$ Find $f(a+h)$ by substituting $x$ with $a+h$: $f(a+h)=\dfrac{a+h}{a+h+1}$ We have $f(a)$ and $f(a+h)$, substitute them in the expression $\dfrac{f(a+h)-f(a)}{h}$ and simplify: $\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{a+h}{a+h+1}-\dfrac{a}{a+1}}{h}=...$ Evaluate the substraction indicated in the numerator and simplify: $...=\dfrac{\dfrac{(a+h)(a+1)-a(a+h+1)}{(a+1)(a+h+1)}}{h}=...$ $...=\dfrac{\dfrac{a^{2}+a+ah+h-a^{2}-ah-a}{(a+1)(a+h+1)}}{h}=\dfrac{\dfrac{h}{(a+1)(a+h+1)}}{h}=...$ $...=\dfrac{1}{(a+1)(a+h+1)}$