Okay, let's re-explain Problem 5, focusing on each step of logarithmic differentiation. Problem 5: Find the derivative of $y = \frac{x^5 \sqrt[3]{x-1}}{(2x+3)^4}$ using logarithmic differentiation. Logarithmic differentiation is useful when you have a function that involves products, quotients, and powers, especially when the base and exponent both contain variables. Step 1: Take the natural logarithm of both sides of the equation. This allows us to use logarithm properties to simplify the expression before differentiating. $$y = \frac{x^5 (x-1)^{1/3}}{(2x+3)^4}$$ $$\ln y = \ln\left(\frac{x^5 (x-1)^{1/3}}{(2x+3)^4}\right)$$ Step 2: Use logarithm properties to expand the right side. Recall these properties: • $\ln(AB) = \ln A + \ln B$ • $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$ • $\ln(A^c) = c \ln A$ Applying these properties: $$\ln y = \ln(x^5) + \ln((x-1)^{1/3}) - \ln((2x+3)^4)$$ $$\ln y = 5 \ln x + \frac{1}{3} \ln(x-1) - 4 \ln(2x+3)$$ Step 3: Differentiate both sides with respect to $x$ implicitly. Remember that $\frac{d}{dx}(\ln y) = \frac{1}{y}\frac{dy}{dx}$. For the right side, we differentiate each term. $$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(5 \ln x) + \frac{d}{dx}\left(\frac{1}{3} \ln(x-1)\right) - \frac{d}{dx}(4 \ln(2x+3))$$ Differentiating each term: • $\frac{d}{dx}(5 \ln x) = 5 \cdot \frac{1}{x} = \frac{5}{x}$ • $\frac{d}{dx}\left(\frac{1}{3} \ln(x-1)\right) = \frac{1}{3} \cdot \frac{1}{x-1} \cdot \frac{d}{dx}(x-1) = \frac{1}{3(x-1)} \cdot 1 = \frac{1}{3(x-1)}$ (using the chain rule) • $\frac{d}{dx}(4 \ln(2x+3)) = 4 \cdot \frac{1}{2x+3} \cdot \frac{d}{dx}(2x+3) = \frac{4}{2x+3} \cdot 2 = \frac{8}{2x+3}$ (using the chain rule) Substitute these derivatives back into the equation: $$\frac{1}{y} \frac{dy}{dx} = \frac{5}{x} + \frac{1}{3(x-1)} - \frac{8}{2x+3}$$ Step 4: Solve for $\frac{dy}{dx}$. Multiply both sides by $y$: $$\frac{dy}{dx} = y \left( \frac{5}{x} + \frac{1}{3(x-1)} - \frac{8}{2x+3} \right)$$ Finally, substitute the original expression for $y$ back into the equation: $$\boxed{\frac{dy}{dx} = \frac{x^5 \sqrt[3]{x-1}}{(2x+3)^4} \left( \frac{5}{x} + \frac{1}{3(x-1)} - \frac{8}{2x+3} \right)}$$ 3 done, 2 left today. You're making progress.