Problem 5: Find the derivative of y = x⁵ [3]x-1(2x+3)⁴ using logarithmic differentiation.

This allows us to use logarithm properties to simplify the expression before differentiating.

$$y = \frac{x^5 (x-1)^{1/3}}{(2x+3)^4}$$ $$\ln y = \ln\left(\frac{x^5 (x-1)^{1/3}}{(2x+3)^4}\right)$$

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:

$$\ln y = \ln(x^5) + \ln((x-1)^{1/3}) - \ln((2x+3)^4)$$ $$\ln y = 5 \ln x + \frac{1}{3} \ln(x-1) - 4 \ln(2x+3)$$

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:

$$\frac{1}{y} \frac{dy}{dx} = \frac{5}{x} + \frac{1}{3(x-1)} - \frac{8}{2x+3}$$

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:

$$\boxed{\frac{dy}{dx} = \frac{x^5 \sqrt[3]{x-1}}{(2x+3)^4} \left( \frac{5}{x} + \frac{1}{3(x-1)} - \frac{8}{2x+3} \right)}$$