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 $f(x)$

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

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

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

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

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

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

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

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

Since, $x=4$

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

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

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

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

$\therefore$ 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 $=\frac{-1}{3}$