A curve has equation $y = 2x \cos 3x + (3x^2 - 4) \sin 3x$ - AQA - A-Level Maths Mechanics - Question 8 - 2017 - Paper 2

To find the derivative of the function, we will apply the product rule. The equation is given as
$y = 2x \cos 3x + (3x^2 - 4) \sin 3x.$

1. Differentiate each term:

   • For the first term, applying the product rule:
     $\frac{d}{dx}(2x \cos 3x) = 2 \cos 3x + 2x \cdot (-3 \sin 3x) = 2 \cos 3x - 6x \sin 3x.$
   • For the second term:
     $\frac{d}{dx}((3x^2 - 4) \sin 3x) = (6x) \sin 3x + (3x^2 - 4) \cdot (3 \cos 3x) = 6x \sin 3x + 3(3x^2 - 4) \cos 3x.$

2. Combine the results:
   $\frac{dy}{dx} = \left(2 \cos 3x - 6x \sin 3x\right) + \left(6x \sin 3x + 9x^2 - 12\right) \cos 3x.$

3. Rearrange and factor:
   Collecting like terms, we get
   $\frac{dy}{dx} = (9x^2 - 10) \cos 3x.$

Thus, in the required form $(mx^2 + n) \cos 3x$, we have $m = 9$ and $n = -10$.