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

Answer

`x+3 y=0`

Explanation

The equation of the given curve is `y=2 x^{2}+3 \sin x` `……. (1)`
Putting `x=0` in `1` , we get, `y=0`.

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

Now,

`y=2 x^{2}+3 \sin x`
$$ \frac{d y}{d x}=4 x+3 \cos x $$
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