1. **State the problem:** Find the derivative of the function $$y = (\sin 2x)^{4x}$$. 2. **Recall the formula:** For a function of the form $$y = [f(x)]^{g(x)}$$, the derivative is found using logarithmic differentiation: $$\ln y = g(x) \ln f(x)$$ Then differentiate both sides: $$\frac{y'}{y} = g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)}$$ So, $$y' = y \left(g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)}\right)$$ 3. **Identify components:** Here, $$f(x) = \sin 2x$$ $$g(x) = 4x$$ 4. **Compute derivatives:** $$f'(x) = 2 \cos 2x$$ (by chain rule) $$g'(x) = 4$$ 5. **Apply the formula:** $$y' = (\sin 2x)^{4x} \left(4 \ln(\sin 2x) + 4x \frac{2 \cos 2x}{\sin 2x}\right)$$ 6. **Simplify the expression:** $$y' = (\sin 2x)^{4x} \left(4 \ln(\sin 2x) + 8x \cot 2x\right)$$ **Final answer:** $$\boxed{y' = (\sin 2x)^{4x} \left(4 \ln(\sin 2x) + 8x \cot 2x\right)}$$ This derivative uses logarithmic differentiation to handle the variable exponent and base.