# 2. Find the slope of the tangent to the curve   $\frac{x-1}{x-2} , x \neq 2 \: \: at\: \: x = 10$

$y = \frac{x-1}{x-2}$
$\left ( \frac{dy}{dx} \right )_{x=10}= \frac{(1)(x-2)-(1)(x-1)}{(x-2)^2} = \frac{x-2-x+1}{(x-2)^2} = \frac{-1}{(x-2)^2}$
$= \frac{-1}{(10-2)^2} = \frac{-1}{8^2} = \frac{-1}{64}$
hence, slope of tangent at x = 10 is $\frac{-1}{64}$