Find the slope of the normal to the curve y = x^3 – x at x=2

Dear Student,

The equation of the curve is $y=x^3-x$. Then,
Slope of the tangent at $x=2$ is given as ${\left[\frac{dy}{dx}\right]}_{x=2}={\left[3x^2-1\right]}_{x=2}=3{\left(2\right)}^2-1=12-1=11$
Slope of normal $=\frac{-1}{slope of \tan gent}=\frac{-1}{11}$
Regards,