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

Solution:

Given:

$y=2 x^{2}+3 \sin x$ at $x=0$

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 $\frac{d y}{d x}$

$\Rightarrow \mathrm{y}=2 \mathrm{x}^{2}+3 \sin \mathrm{x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \times \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+3 \times \frac{\mathrm{dy}}{\mathrm{dx}}(\sin \mathrm{x})$

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

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})=\cos \mathrm{x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4 \mathrm{x}+3 \cos \mathrm{x}$

Since, $x=2$

$\therefore \cos (0)=1$

$\Rightarrow\left(\frac{d y}{d x}\right) x=0=0+3 x 1$

$\Rightarrow\left(\frac{d y}{d x}\right) x=0=3$

$\therefore$ The Slope of the tangent at $x=0$ is 3

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

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \mathrm{x}=0}$

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