Find the slope of the tangent to curve $y = {x^3} - x + 1$ at the point whose $x -$coordinate is $2$.

We have, $y = {x^3} - x + 1$ …(i)

Differentiating (i) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = 3{x^2} - 1$

Therefore, Slope of tangent at $x = 2$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{x = 2}} = 3{(2)^2} - 1 = 11$