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

Given:

$y=\sqrt{x^{3}}$ at $x=4$

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

$y=\sqrt{x^{3}}$

$\therefore \sqrt[n]{x}=x^{\frac{1}{n}}$

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

$\Rightarrow y=(x)^{\frac{3}{2}}$

$\therefore \frac{d y}{d x}\left(x^{n}\right)=n \cdot x^{n-1}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3}{2}(\mathrm{x})^{\frac{3}{2}-1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3}{2}(\mathrm{x})^{\frac{1}{2}}$

Since, $\mathrm{x}=4$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \mathrm{x}=4=\frac{3}{2}(4)^{\frac{1}{2}}$

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

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

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

$\therefore$ The Slope of the tangent at $x=4$ is 3

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

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right) x=4}$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{3}$