1. **Problem (a):** Evaluate $\frac{dy}{dx}$ at $x=2.5$ for $y=\frac{2x^2+3}{\ln(2x)}$. 2. **Formula and rules:** Use the quotient rule for derivatives: if $y=\frac{u}{v}$, then $$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$ where $u=2x^2+3$ and $v=\ln(2x)$. 3. **Calculate derivatives:** - $\frac{du}{dx} = 4x$ - $\frac{dv}{dx} = \frac{1}{2x} \times 2 = \frac{1}{x}$ (chain rule on $\ln(2x)$) 4. **Apply quotient rule:** $$\frac{dy}{dx} = \frac{\ln(2x) \cdot 4x - (2x^2+3) \cdot \frac{1}{x}}{(\ln(2x))^2}$$ 5. **Simplify numerator:** $$4x \ln(2x) - \frac{2x^2+3}{x} = 4x \ln(2x) - 2x - \frac{3}{x}$$ 6. **Evaluate at $x=2.5$:** - $\ln(2 \times 2.5) = \ln(5) \approx 1.60944$ - Numerator: $4 \times 2.5 \times 1.60944 - 2 \times 2.5 - \frac{3}{2.5} = 16.0944 - 5 - 1.2 = 9.8944$ - Denominator: $(1.60944)^2 = 2.5903$ 7. **Final value:** $$\frac{dy}{dx} = \frac{9.8944}{2.5903} \approx 3.82$$ (3 significant figures) --- **Answer (a):** $\boxed{3.82}$ --- **Slug:** derivative evaluation **Subject:** calculus **Desmos:** {"latex":"y=\frac{2x^2+3}{\ln(2x)}","features":{"intercepts":true,"extrema":true}} **q_count:** 4