Find the slope of the tangent to the curve $y = \dfrac{x-1}{x-2}$, $x \ne 2$ and $x=10$

The correct answer is $\dfrac{-1}{64}$

Given $y = \dfrac{x-1}{x-2}$

∴ $\dfrac{dy}{dx} = \dfrac{(x-2)(1) - (x-1)(1)}{(x-2)^2}$ = $\dfrac{-1}{(x-2)^2}$


∴ slope = $\left( \dfrac{dy}{dx} \right)_{x=10} $ = $\dfrac{-1}{(10-2)^2}$ = $\dfrac{-1}{64}$