Solution:
To differentiate the given function $$ f(x) = \frac{x^3 (3 - 2x)^2}{\sqrt[3]{x} + 7} $$ using logarithmic differentiation, we first take the natural logarithm of both sides of the equation defining $$ f(x) $$:1. Take the natural logarithm of f(x):\[ \ln(f(x)) = \ln\left(\frac{x^3 (3 - 2x)^2}{\sqrt[3]{x} + 7}\right) \]2. Apply properties of logarithms to simplify the right-hand side:\[ \ln(f(x)) = \ln(x^3) + \ln((3 - 2x)^2) - \ln(\sqrt[3]{x} + 7) \]\[ \ln(f(x)) = 3\ln(x) + 2\ln(3 - 2x) - \ln(\sqrt[3]{x} + 7) \]3. Differentiate both sides with respect to x:\[ \frac{1}{f(x)} \cdot f'(x) = \frac{d}{dx}(3\ln(x)) + \frac{d}{dx}(2\ln(3 - 2x)) - \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) \]4. Apply the derivatives;- For $$ \frac{d}{dx}(3\ln(x)) $$, use the derivative of $$\ln(x)$$, which is $$ \frac{1}{x} $$:\[ \frac{d}{dx}(3\ln(x)) = 3 \cdot \frac{1}{x} = \frac{3}{x} \]- For $$ \frac{d}{dx}(2\ln(3 - 2x)) $$, use the chain rule:\[ \frac{d}{dx}(2\ln(3 - 2x)) = 2 \cdot \frac{1}{(3 - 2x)} \cdot (-2) = \frac{-4}{(3 - 2x)} \]- For $$ \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) $$, again use the chain rule:\[ \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) = \frac{1}{(\sqrt[3]{x} + 7)} \cdot \frac{d}{dx}(\sqrt[3]{x}) \]\[ \frac{d}{dx}(\sqrt[3]{x}) = \frac{1}{3x^{2/3}} \]So,\[ \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) = \frac{1}{(\sqrt[3]{x} + 7)} \cdot \frac{1}{3x^{2/3}} \]5. Now combine these results:\[ \frac{1}{f(x)} \cdot f'(x) = \frac{3}{x} - \frac{4}{3 - 2x} - \frac{1}{3x^{2/3}(\sqrt[3]{x} + 7)} \]6. Solve for $$ f'(x) $$:\[ f'(x) = f(x) \cdot \left( \frac{3}{x} - \frac{4}{3 - 2x} - \frac{1}{3x^{2/3}(\sqrt[3]{x} + 7)} \right) \]7. Replace $$ f(x) $$ with the original function:\[ f'(x) = \frac{x^3 (3 - 2x)^2}{\sqrt[3]{x} + 7} \cdot \left( \frac{3}{x} - \frac{4}{3 - 2x} - \frac{1}{3x^{2/3}(\sqrt[3]{x} + 7)} \right) \]This is the derivative of the function using logarithmic differentiation. You can simplify the expression further by combining terms and multiplying through by the original function.