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

Solution:

Given:

$y=\sqrt{x}$ at $x=9$

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

$\Rightarrow y=\sqrt{x}$

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

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

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

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

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

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

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

Since, $x=9$

$\left(\frac{d y}{d x}\right) x=9=\frac{1}{2}(9)^{\frac{-1}{2}}$

$\Rightarrow\left(\frac{d y}{d x}\right) x=9=\frac{1}{2} \times \frac{1}{(9)^{\frac{1}{2}}}$

$\Rightarrow\left(\frac{d y}{d x}\right) x=9=\frac{1}{2} \times \frac{1}{\sqrt{9}}$

$\Rightarrow\left(\frac{d y}{d x}\right) x=9=\frac{1}{2} \times \frac{1}{3}$

$\Rightarrow\left(\frac{d y}{d x}\right) x=9=\frac{1}{6}$

$\therefore$ The Slope of the tangent at $x=9$ is $\frac{1}{6}$

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

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right) \mathrm{x}=9}$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\frac{1}{6}}$

$\Rightarrow$ The Slope of the normal $=-6$