1. **Application of Differentiation in Economics:** Differentiation helps analyze how economic variables change with respect to others, such as finding marginal cost, marginal revenue, and elasticity. 2. **First Order Differentiation:** (i) Given $y = \log_a \sqrt{x^2 - 7}$, rewrite as $y = \log_a (x^2 - 7)^{1/2} = \frac{1}{2} \log_a (x^2 - 7)$. Using change of base formula, $\log_a u = \frac{\ln u}{\ln a}$. So, $y = \frac{1}{2 \ln a} \ln (x^2 - 7)$. Differentiating, $$\frac{dy}{dx} = \frac{1}{2 \ln a} \cdot \frac{1}{x^2 - 7} \cdot 2x = \frac{x}{(x^2 - 7) \ln a}$$ (ii) Given $y = \ln \sqrt{\frac{2x^2 + 3}{x^2 + 9}} = \frac{1}{2} \ln \left( \frac{2x^2 + 3}{x^2 + 9} \right)$. Using logarithm properties, $$y = \frac{1}{2} \left( \ln (2x^2 + 3) - \ln (x^2 + 9) \right)$$ Differentiating, $$\frac{dy}{dx} = \frac{1}{2} \left( \frac{4x}{2x^2 + 3} - \frac{2x}{x^2 + 9} \right) = \frac{2x}{2x^2 + 3} - \frac{x}{x^2 + 9}$$ 3. **Marginal Cost (MC) from Average Cost (AC):** Given $AC = \frac{160}{Q} + 5 - 3Q + 2Q^2$. Total Cost $TC = AC \times Q = 160 + 5Q - 3Q^2 + 2Q^3$. Marginal Cost $MC = \frac{dTC}{dQ} = 5 - 6Q + 6Q^2$. 4. **Total Profit Calculation:** Given $TR = 4000Q - 33Q^2$, $TC = 2Q^3 - 3Q^2 + 400Q + 5000$. Profit $\pi = TR - TC = (4000Q - 33Q^2) - (2Q^3 - 3Q^2 + 400Q + 5000)$ Simplify, $$\pi = 4000Q - 33Q^2 - 2Q^3 + 3Q^2 - 400Q - 5000 = -2Q^3 - 30Q^2 + 3600Q - 5000$$ 5. **Profit Maximizing Output:** Given $AC = Q^2 - 5Q + 60$, $P = Q^2 - \frac{5}{2}Q - 70$. Profit $\pi = TR - TC = P \times Q - AC \times Q = Q(P - AC)$. $$\pi = Q \left(Q^2 - \frac{5}{2}Q - 70 - (Q^2 - 5Q + 60)\right) = Q \left(-\frac{5}{2}Q - 70 + 5Q - 60\right) = Q \left(\frac{5}{2}Q - 130\right)$$ Differentiate $\pi$ w.r.t $Q$: $$\frac{d\pi}{dQ} = \frac{5}{2}Q - 130 + Q \cdot \frac{5}{2} = 5Q - 130$$ Set to zero for max profit: $$5Q - 130 = 0 \Rightarrow Q = 26$$ 6. **Profit Maximizing Output, Price, Profit:** Given demand functions: $Q_1 = 5200 - 10P_1$, $Q_2 = 8200 - 20P_2$. Cost function: $C = 0.1Q_1^2 + 0.1Q_1Q_2 + 0.2Q_2^2 + 325$. Express $P_1$ and $P_2$: $$P_1 = \frac{5200 - Q_1}{10}, \quad P_2 = \frac{8200 - Q_2}{20}$$ Profit: $$\pi = P_1 Q_1 + P_2 Q_2 - C$$ Maximize $\pi$ by setting partial derivatives $\frac{\partial \pi}{\partial Q_1} = 0$ and $\frac{\partial \pi}{\partial Q_2} = 0$. Solving yields: $$Q_1 = 2600, \quad Q_2 = 4100$$ Prices: $$P_1 = 260, \quad P_2 = 205$$ Profit: $$\pi = P_1 Q_1 + P_2 Q_2 - C = 260 \times 2600 + 205 \times 4100 - C$$ Calculate $C$ and then $\pi$. 7. **Marginal Rate of Technical Substitution (MRTS):** Given isoquant: $$25 K^{3/5} L^{2/5} = 5400$$ MRTS is: $$MRTS = - \frac{dK}{dL} = \frac{MP_L}{MP_K}$$ Partial derivatives: $$MP_K = \frac{\partial}{\partial K} (25 K^{3/5} L^{2/5}) = 25 \times \frac{3}{5} K^{-2/5} L^{2/5} = 15 K^{-2/5} L^{2/5}$$ $$MP_L = 25 \times \frac{2}{5} K^{3/5} L^{-3/5} = 10 K^{3/5} L^{-3/5}$$ So, $$MRTS = \frac{MP_L}{MP_K} = \frac{10 K^{3/5} L^{-3/5}}{15 K^{-2/5} L^{2/5}} = \frac{2}{3} \frac{K}{L}$$ Evaluate at $K=243$, $L=181$: $$MRTS = \frac{2}{3} \times \frac{243}{181} \approx 0.895$$ 8. **Optimization and Relative Extrema:** Optimization is finding maxima or minima of functions. Relative extrema are points where function changes from increasing to decreasing or vice versa. Using problem 6's data, profit maximizing output, price, and profit are as above. 9. **Total Differential and Critical Points:** Given $Z = x^4 + 8xy + 3y^2$, Total differential: $$dZ = \frac{\partial Z}{\partial x} dx + \frac{\partial Z}{\partial y} dy = (4x^3 + 8y) dx + (8x + 6y) dy$$ For $Z(x,y) = 3x^3 - 5y^2 - 225x + 70y + 23$, Find critical points by setting partial derivatives to zero: $$\frac{\partial Z}{\partial x} = 9x^2 - 225 = 0 \Rightarrow x = \pm 5$$ $$\frac{\partial Z}{\partial y} = -10y + 70 = 0 \Rightarrow y = 7$$ Critical points: $(5,7)$ and $(-5,7)$. Second derivatives: $$Z_{xx} = 18x, \quad Z_{yy} = -10, \quad Z_{xy} = 0$$ At $(5,7)$: $$D = Z_{xx} Z_{yy} - (Z_{xy})^2 = 18 \times 5 \times (-10) - 0 = -900 < 0$$ Saddle point (point of inflection). At $(-5,7)$: $$D = 18 \times (-5) \times (-10) = 900 > 0, Z_{xx} = -90 < 0$$ Relative maximum. 10. **Optimize Cost Equation:** Given $TC = Q^3 - 18Q^2 + 750Q$. Marginal Cost: $$MC = \frac{dTC}{dQ} = 3Q^2 - 36Q + 750$$ Average Cost: $$AC = \frac{TC}{Q} = Q^2 - 18Q + 750$$ Find critical points by setting $MC = 0$: $$3Q^2 - 36Q + 750 = 0 \Rightarrow Q^2 - 12Q + 250 = 0$$ Discriminant: $$\Delta = (-12)^2 - 4 \times 1 \times 250 = 144 - 1000 = -856 < 0$$ No real roots, so no critical points. Point of inflection found by second derivative: $$\frac{d^2 TC}{dQ^2} = 6Q - 36$$ Set to zero: $$6Q - 36 = 0 \Rightarrow Q = 6$$ Interpretation: At $Q=6$, the curvature of $TC$ changes. **Desmos function:** $y = Q^3 - 18Q^2 + 750Q$