Find the The Slopes of the tangent and the normal to the following curves at the indicated points:

Solution:

Given:

$y=x^{3}-x$ at $x=2$

First, we have to find $\frac{d y}{d x}$ of given function, $f(x)$, i.e, to find the derivative of $f(x)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}$

The Slope of the tangent is $\frac{d y}{d x}$

$\Rightarrow y=x^{3}-x$

$\Rightarrow \frac{d y}{d x}=\frac{d y}{d x}\left(x^{3}\right)+3 x \frac{d y}{d x}(x)$

$\Rightarrow \frac{d y}{d x}=3 \cdot x^{3}-1-1 \cdot x^{1-0}$

$\Rightarrow \frac{d y}{d x}=3 x^{2}-1$

Since, $x=2$

$\Rightarrow\left(\frac{d y}{d x}\right) x=2=3 x(2)^{2}-1$

$\Rightarrow\left(\frac{d y}{d x}\right) x=2=(3 \times 4)-1$

$\Rightarrow\left(\frac{d y}{d x}\right) x=2=12-1$

$\Rightarrow\left(\frac{d y}{d x}\right) x=2=11$

$\therefore$ The Slope of the tangent at $x=2$ is 11

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\text { The Slope of the tangent }}$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \mathrm{x}=2}$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{11}$