Answer
$-\dfrac{1}{(a+1)(a+h+1)}$
Work Step by Step
$f(x)=\dfrac{1}{x+1}$
To find $f(a)$, substitute $x$ with $a$:
$f(a)=\dfrac{1}{a+1}$
To find $f(a+h)$, substitute $x$ with $a+h$:
$f(a+h)=\dfrac{1}{(a+h)+1}$
Since we have $f(a)$ and $f(a+h)$, substitute into the expression $\dfrac{f(a+h)-f(a)}{h}:$
$\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{1}{a+h+1}-\dfrac{1}{a+1}}{h}=...$
Simplify:
$...=\dfrac{\dfrac{(a+1)-(a+h+1)}{(a+1)(a+h+1)}}{h}=\dfrac{\dfrac{a+1-a-h-1}{(a+1)(a+h+1)}}{h}=...$
$...=\dfrac{\dfrac{-h}{(a+1)(a+h+1)}}{h}=-\dfrac{1}{(a+1)(a+h+1)}$
