## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{\dfrac{1}{1+x+h}-\dfrac{1}{1+x}}{h}=-\dfrac{1}{(1+x+h)(1+x)}$
$\dfrac{\dfrac{1}{1+x+h}-\dfrac{1}{1+x}}{h}$ 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: $...=-\dfrac{1}{(1+x+h)(1+x)}$