# Solutions To Mathematics Textbooks/Calculus Early Transcendentals (6th) (0495011665)/Chapter 1.1

Replace instances of ${\displaystyle f(x)\,}$ with the expression ${\displaystyle {\frac {1}{x}}}$.
${\displaystyle {\frac {{\frac {1}{x}}-{\frac {1}{a}}}{x-a}}=\left({\frac {1}{x}}-{\frac {1}{a}}\right)\cdot {\frac {1}{x-a}}}$
${\displaystyle ={\frac {a-x}{xa(x-a)}}={\frac {(a-x)}{-xa(a-x)}}=-{\frac {1}{ax}}}$