Question

Find the slope of the tangent and the normal to the following curves at the indicated points.
$y=x^3-x$ at $x=2$.

Answer 1

The equation of the curve is $y=x^3-x$.
$\therefore \frac{dy}{dx}=3x^2-1$
At $x=2$, we get: $\frac{dy}{dx}=3\times 2^2-1=11$
Hence, slope of the tangent at $x=2$ is $11$ and that of the normal is $-\frac{1}{11}$.