Answer
$\dfrac{\dfrac{1}{1+x+h}-\dfrac{1}{1+x}}{h}=-\dfrac{1}{(1+x+h)(1+x)}$
Work Step by Step
$\dfrac{\dfrac{1}{1+x+h}-\dfrac{1}{1+x}}{h}$
Evaluate the difference of fractions in the numerator:
$\dfrac{\dfrac{1}{1+x+h}-\dfrac{1}{1+x}}{h}=\dfrac{\dfrac{(1+x)-(1+x+h)}{(1+x+h)(1+x)}}{h}=...$
$...=\dfrac{\dfrac{1+x-1-x-h}{(1+x+h)(1+x)}}{h}=\dfrac{\dfrac{-h}{(1+x+h)(1+x)}}{h}=...$
Evaluate the division and simplify if possible:
$...=-\dfrac{1}{(1+x+h)(1+x)}$
