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The slope of normal to the curve y=3x2+3sin⁡3xy = 3x^2 + 3 \sin 3x, at x=0x = 0 is:

Solution

✅ Correct Option: 3

dydx=6x+9cos⁡3x\frac{dy}{dx} = 6x + 9\cos 3x.

At x=0x = 0: slope of tangent = 0+9=90 + 9 = 9.

Slope of normal = −19-\frac{1}{9}.

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