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

We have, $y = {x^2} - 3x + 2$ …(i)

Differentiating (i) w.r.t $x$, we get $\cfrac{{dy}}{{dx}} = 3{x^2} - 3$
Therefore the slope of tangent at $x = 3$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{x = 3}} = 3 \times {3^2} - 3 = 24$.