To find \(\frac{dy}{dx}\) for the given function \(y = 2x \cos(2x) - \sin(2x)\), we need to differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{d}{dx}(2x \cos(2x) - \sin(2x))\)

Using the product rule for differentiation, which states that \(\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}\), where \(u\) and \(v\) are functions of \(x\), and \(\frac{du}{dx}\) and \(\frac{dv}{dx}\) are their derivatives, we differentiate each term of the given function:

\(u = 2x, \quad \frac{du}{dx} = 2\)

\(v = \cos(2x), \quad \frac{dv}{dx} = -2 \sin(2x)\)

\(w = -\sin(2x), \quad \frac{dw}{dx} = -2 \cos(2x)\)

Applying the product rule:

\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} + \frac{dw}{dx}\)

\(\frac{dy}{dx} = 2x \cdot (-2 \sin(2x)) + \cos(2x) \cdot 2 + (-2 \cos(2x))\)

Simplify:

\(\frac{dy}{dx} = -4x \sin(2x) + 2 \cos(2x) - 2 \cos(2x)\)

\(\frac{dy}{dx} = -4x \sin(2x)\)

Now, we want to find \(\frac{dy}{dx}\) when \(x = \frac{\pi}{4}\):

\(\frac{dy}{dx} \Big|_{x = \frac{\pi}{4}} = -4 \cdot \frac{\pi}{4} \sin\left(2 \cdot \frac{\pi}{4}\right)\)

Since \(\sin\left(\frac{\pi}{2}\right) = 1\), we have:

\(\frac{dy}{dx} \Big|_{x = \frac{\pi}{4}} = -\pi \cdot 1 = -\pi\)

Therefore, the value of \(\frac{dy}{dx}\) when \(x = \frac{\pi}{4}\) is \(-\pi\).