1. **State the problem:** Find the derivative of the function $$y = x^{3x}$$ using logarithmic differentiation. 2. **Recall the formula and rules:** For functions of the form $$y = f(x)^{g(x)}$$, logarithmic differentiation is useful. We take the natural logarithm of both sides: $$\ln y = g(x) \ln f(x)$$ Then differentiate implicitly using the product rule and chain rule. 3. **Apply logarithmic differentiation:** Take natural log of both sides: $$\ln y = 3x \ln x$$ 4. **Differentiate both sides with respect to $$x$$:** Using implicit differentiation: $$\frac{1}{y} \frac{dy}{dx} = 3 \ln x + 3x \cdot \frac{1}{x} = 3 \ln x + 3$$ 5. **Solve for $$\frac{dy}{dx}$$:** Multiply both sides by $$y$$: $$\frac{dy}{dx} = y (3 \ln x + 3)$$ Recall that $$y = x^{3x}$$, so: $$\frac{dy}{dx} = x^{3x} (3 \ln x + 3) = 3 x^{3x} (\ln x + 1)$$ **Final answer:** $$y' = 3 x^{3x} (\ln x + 1)$$