Explanation

(d)
Let $y = 2\cos{x\cos{3x}}$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = 2\cos{x( - 3\sin{3x) + 2\cos{3x( - \sin{x)}}}}$
$= - 3(\sin{4x + \sin{2x) + ( - 1)\lbrack\sin{4x + \sin{( - 2x)\rbrack}}}}$
Again differentiating w.r.t. $x$, we get
$\frac{d^{2}y}{dx^{2}} = - 3(4\cos{4x + 2\cos{2x) - 1(4\cos{4x - 2\cos{2x}}}})$
$= - 16\cos{4x - 4\cos{2x = - 4(\cos{2x + 4\cos{4x)}}}}$
$= - 2^{2}(\cos{2x + 2^{2}\cos{4x)}}$