Find the equation of the normal to the curve `y=2x^2+3sinxa tx=0.`

Answer

`x+3 y=0`

Explanation

Given the equation of the curve is `y=2 x^{2}+3 \sin x \ldots . .[1]`


Putting `x=0` in [1], we get `y=0`.


So, the point is `(0,0)`.


Now, `y=2 x^{2}+3 \sin x`


$$
\begin{aligned}
&\Rightarrow \frac{d y}{d x}=4 x+3 \cos x \\
&\Rightarrow\left(\frac{d y}{d x}\right)_{(0,0)}=3 .
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