1. Problem (a): Evaluate $\frac{dy}{dx}$ at $x=2.5$ for $y=\frac{2x^2+3}{\ln(2x)}$. Formula: Use the quotient rule for differentiation: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$ where $u=2x^2+3$ and $v=\ln(2x)$. Step 1: Differentiate $u$: $$\frac{du}{dx} = 4x$$ Step 2: Differentiate $v$: $$\frac{dv}{dx} = \frac{1}{2x} \times 2 = \frac{1}{x}$$ Step 3: Apply quotient rule: $$\frac{dy}{dx} = \frac{\ln(2x) \cdot 4x - (2x^2+3) \cdot \frac{1}{x}}{(\ln(2x))^2}$$ Step 4: Simplify numerator: $$4x \ln(2x) - \frac{2x^2+3}{x} = 4x \ln(2x) - 2x - \frac{3}{x}$$ Step 5: Evaluate at $x=2.5$: Calculate $\ln(2 \times 2.5) = \ln(5) \approx 1.60944$ Numerator: $$4 \times 2.5 \times 1.60944 - 2 \times 2.5 - \frac{3}{2.5} = 16.0944 - 5 - 1.2 = 9.8944$$ Denominator: $$(1.60944)^2 = 2.5903$$ Therefore, $$\frac{dy}{dx} = \frac{9.8944}{2.5903} \approx 3.82$$ --- 2. Problem (b): Find $\frac{dy}{dx}$ for $$y = \frac{1}{(x^3 - 2x + 1)^5} = (x^3 - 2x + 1)^{-5}$$ Formula: Use the chain rule: $$\frac{dy}{dx} = -5 (x^3 - 2x + 1)^{-6} \cdot \frac{d}{dx}(x^3 - 2x + 1)$$ Step 1: Differentiate inner function: $$\frac{d}{dx}(x^3 - 2x + 1) = 3x^2 - 2$$ Step 2: Substitute: $$\frac{dy}{dx} = -5 (x^3 - 2x + 1)^{-6} (3x^2 - 2)$$ --- 3. Problem (c): Find the second derivative $\frac{d^2y}{dx^2}$ for $$y = (2x - 3)^4$$ Step 1: First derivative using chain rule: $$\frac{dy}{dx} = 4(2x - 3)^3 \cdot 2 = 8(2x - 3)^3$$ Step 2: Second derivative: $$\frac{d^2y}{dx^2} = 8 \cdot 3 (2x - 3)^2 \cdot 2 = 48 (2x - 3)^2$$ --- 4. Problem (d): Given alternating current $$I = 10 \sin(2 \pi f t)$$ with $f=150$ Hz and $t=20$ ms = 0.02 s. Step 1: Differentiate $I$ with respect to $t$: $$\frac{dI}{dt} = 10 \cdot 2 \pi f \cos(2 \pi f t) = 20 \pi f \cos(2 \pi f t)$$ Step 2: Substitute values: $$\frac{dI}{dt} = 20 \pi \times 150 \times \cos(2 \pi \times 150 \times 0.02)$$ Calculate inside cosine: $$2 \pi \times 150 \times 0.02 = 6 \pi$$ Since $\cos(6 \pi) = 1$, then $$\frac{dI}{dt} = 20 \pi \times 150 \times 1 = 3000 \pi \approx 9424.78$$ Final answer: Rate of change of current at $t=20$ ms is approximately 9425 amperes per second.