Let $y = (5x)^{3 \cos 2x}$.
Taking the natural logarithm on both sides,we get:
$\ln y = 3 \cos 2x \cdot \ln(5x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = 3 \left[ \ln(5x) \cdot \frac{d}{dx}(\cos 2x) + \cos 2x \cdot \frac{d}{dx}(\ln 5x) \right]$.
Applying the chain rule:
$\frac{1}{y} \frac{dy}{dx} = 3 \left[ \ln(5x) \cdot (-2 \sin 2x) + \cos 2x \cdot \frac{1}{5x} \cdot 5 \right]$.
Simplifying the expression:
$\frac{1}{y} \frac{dy}{dx} = 3 \left[ -2 \sin 2x \ln(5x) + \frac{\cos 2x}{x} \right]$.
Multiplying by $y$:
$\frac{dy}{dx} = 3(5x)^{3 \cos 2x} \left[ \frac{\cos 2x}{x} - 2 \sin 2x \ln(5x) \right]$.
Thus,$\frac{dy}{dx} = (5x)^{3 \cos 2x} \left[ \frac{3 \cos 2x}{x} - 6 \sin 2x \ln(5x) \right]$.