The equation of the normal to the curve $y=4x^3+2sin x $ at (0, 3) is :

$x-3y =3$ $x+2y =6$ $2x-y=-3$ $2x+y = 3$

$x+2y =6$

The correct answer is Option (2) → $x+2y =6$ $y=4x^3+2\sin x$  P(0, 3) $\frac{dy}{dx}=12x^2+2\cos x$ Slope of normal is $-\frac{dx}{dy}$ $-\frac{dx}{dy}=-\frac{1}{12x^2+2\cos x}$ so $\left.-\frac{dx}{dy}\right]_{x=0}=-\frac{1}{2}$ eq. of normal → $y-3=-\frac{1}{2}(x-0)$ so $2y-6=-x⇒x+2y=6$