Find the slope of the tangent to the curve $y = x^3 - x$ at $x=2$

The correct answer is 11

Given $y = x^3 - x$

∴ $\dfrac{dy}{dx}$ = $3x^2 - 1$

∴ slope of the tangent at $x=2$ is

$\dfrac{dy}{dx}$ = $3x^2 - 1$ = $3 (2)^2 - 1$ = $11$