Answer
$f(a)=\frac{a}{a+1}$
$f(a+h)=\frac{a+h}{a+h+1}$
$\frac{f(a+h)-f(a)}{h}=\frac{1}{(a+h+1)(a+1)}$
Work Step by Step
We are given $f(x)=\frac{x}{x+1}$
We evaluate:
$f(a)=\frac{a}{a+1}$
$f(a+h)=\frac{a+h}{a+h+1}$
$\frac{f(a+h)-f(a)}{h}=\frac{\frac{a+h}{a+h+1}-\frac{a}{a+1}}{h}=\frac{(a+h)(a+1)-a(a+h+1)}{(a+h+1)(a+1)h}=\frac{a^{2}+a+ah+h-a^{2}-ah-a}{(a+h+1)(a+1)h}=\frac{h}{(a+h+1)(a+1)h}=\frac{1}{(a+h+1)(a+1)}$