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

We have, $y = \cfrac{{x - 1}}{{x - 2}},x \ne 2$ …(i)

Differentiating (i) w.r.t. $x$, we get
$\cfrac{{dy}}{{dx}} = \cfrac{{(x - 2) \cdot 1 - (x - 1) \cdot 1}}{{{{(x - 2)}^2}}} = \cfrac{{ - 1}}{{{{(x - 2)}^2}}}$

Therefore the slope of tangent at $x = 10$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{x = 10}} = \cfrac{{ - 1}}{{{{(10 - 2)}^2}}} = - \cfrac{1}{{64}}$