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