1. Problem: Find the derivatives of the given functions. (a) Given $y = \frac{9 - 15x + 4^3 - 7x^5}{x^4}$. Step 1: Simplify numerator: $4^3 = 64$, so numerator is $9 - 15x + 64 - 7x^5 = 73 - 15x - 7x^5$. Step 2: Rewrite $y = \frac{73 - 15x - 7x^5}{x^4} = 73x^{-4} - 15x^{-3} - 7x^{1}$. Step 3: Differentiate term-by-term using power rule $\frac{d}{dx} x^n = nx^{n-1}$: $$y' = 73(-4)x^{-5} - 15(-3)x^{-4} - 7(1)x^{0} = -292x^{-5} + 45x^{-4} - 7$$ (b) Given $f(t) = -4 \sqrt[5]{t^3 - 1} \cdot \frac{3}{\sqrt{t^2 + 1}}$. Step 1: Rewrite as $f(t) = -12 (t^3 - 1)^{1/5} (t^2 + 1)^{-1/2}$. Step 2: Use product rule: $f' = -12 \left[ \frac{1}{5}(t^3 - 1)^{-4/5} \cdot 3t^2 \cdot (t^2 + 1)^{-1/2} + (t^3 - 1)^{1/5} \cdot \left(-\frac{1}{2}\right)(t^2 + 1)^{-3/2} \cdot 2t \right]$. Step 3: Simplify: $$f'(t) = -12 \left[ \frac{3t^2}{5}(t^3 - 1)^{-4/5}(t^2 + 1)^{-1/2} - t (t^3 - 1)^{1/5} (t^2 + 1)^{-3/2} \right]$$ Step 4: Evaluate at $t=8$ by substituting values. (c) Given $f(x) = \frac{x^3 + 2x^2 + 3}{(5x - 2)(3x^2 + 2)}$. Step 1: Let $u = x^3 + 2x^2 + 3$, $v = (5x - 2)(3x^2 + 2)$. Step 2: Use quotient rule: $f' = \frac{u'v - uv'}{v^2}$. Step 3: Compute derivatives: $u' = 3x^2 + 4x$, $v = (5x - 2)(3x^2 + 2)$, $v' = 5(3x^2 + 2) + (5x - 2)(6x) = 15x^2 + 10 + 30x^2 - 12x = 45x^2 - 12x + 10$. Step 4: Substitute and simplify. (d) Given $y = (3\sqrt{x} - 5)(2\sqrt{x} + x^3 + 7)$. Step 1: Use product rule: $y' = (3\sqrt{x} - 5)'(2\sqrt{x} + x^3 + 7) + (3\sqrt{x} - 5)(2\sqrt{x} + x^3 + 7)'$. Step 2: Compute derivatives: $(3\sqrt{x} - 5)' = 3 \cdot \frac{1}{2} x^{-1/2} = \frac{3}{2} x^{-1/2}$, $(2\sqrt{x} + x^3 + 7)' = 2 \cdot \frac{1}{2} x^{-1/2} + 3x^2 + 0 = x^{-1/2} + 3x^2$. Step 3: Substitute and simplify. (e) Given $y = 5 + 2x - 7x^2 (6x^3 - x^2 + 3x + 5)^2$. Step 1: Use sum and product rules. Step 2: Derivative of $5 + 2x$ is $2$. Step 3: For $-7x^2 (g(x))^2$, use product rule: $u = -7x^2$, $v = (g(x))^2$, where $g(x) = 6x^3 - x^2 + 3x + 5$. Step 4: $u' = -14x$, $v' = 2g(x) g'(x)$. Step 5: Compute $g'(x) = 18x^2 - 2x + 3$. Step 6: Apply product rule: $$y' = 2 + u'v + uv' = 2 - 14x (g(x))^2 - 7x^2 \cdot 2 g(x) g'(x)$$ (f) Given $y = \sqrt[5]{\frac{8x^2 - 5}{x^2 - 3}}$. Step 1: Rewrite as $y = \left( \frac{8x^2 - 5}{x^2 - 3} \right)^{1/5}$. Step 2: Use chain rule: $$y' = \frac{1}{5} \left( \frac{8x^2 - 5}{x^2 - 3} \right)^{-4/5} \cdot \frac{d}{dx} \left( \frac{8x^2 - 5}{x^2 - 3} \right)$$ Step 3: Use quotient rule for inner derivative: $$\frac{d}{dx} \left( \frac{8x^2 - 5}{x^2 - 3} \right) = \frac{(16x)(x^2 - 3) - (8x^2 - 5)(2x)}{(x^2 - 3)^2}$$ (g) Given $y = \ln \left( \sqrt{ \frac{x^4 + 2x^2 + 10}{x^4 + 4} } \right)$. Step 1: Simplify inside log: $$y = \ln \left( \left( \frac{x^4 + 2x^2 + 10}{x^4 + 4} \right)^{1/2} \right) = \frac{1}{2} \ln \left( \frac{x^4 + 2x^2 + 10}{x^4 + 4} \right)$$ Step 2: Use log properties: $$y = \frac{1}{2} \left( \ln(x^4 + 2x^2 + 10) - \ln(x^4 + 4) \right)$$ Step 3: Differentiate: $$y' = \frac{1}{2} \left( \frac{4x^3 + 4x}{x^4 + 2x^2 + 10} - \frac{4x^3}{x^4 + 4} \right)$$ (h) Given $y = e^{3x^3 + 9x^2 + 2x + 1}$. Step 1: Use chain rule: $$y' = e^{3x^3 + 9x^2 + 2x + 1} \cdot (9x^2 + 18x + 2)$$ (i) Given $y = \ln \left[ ((x^2 + 4)(x^3 + 8x - 3))^2 \right]$. Step 1: Use log power rule: $$y = 2 \ln \left( (x^2 + 4)(x^3 + 8x - 3) \right) = 2 \left( \ln(x^2 + 4) + \ln(x^3 + 8x - 3) \right)$$ Step 2: Differentiate: $$y' = 2 \left( \frac{2x}{x^2 + 4} + \frac{3x^2 + 8}{x^3 + 8x - 3} \right)$$ (j) Given $f(x) = 5x^2 \ln x$. Step 1: Use product rule: $$f'(x) = 5 \left( 2x \ln x + x^2 \cdot \frac{1}{x} \right) = 5 (2x \ln x + x) = 5x (2 \ln x + 1)$$ Step 2: Evaluate at $x=1$: $$f'(1) = 5 \cdot 1 (2 \cdot 0 + 1) = 5$$ Step 3: Evaluate at $x=5$: $$f'(5) = 5 \cdot 5 (2 \ln 5 + 1) = 25 (2 \ln 5 + 1)$$ --- 7. Find $\frac{dy}{dx}$, $\frac{d^2 y}{dx^2}$, and $\frac{d^3 y}{dx^3}$. (a) Given $\ln(xy) + x = 4x^2$. Step 1: Differentiate implicitly: $$\frac{1}{xy} (y + x \frac{dy}{dx}) + 1 = 8x$$ Step 2: Multiply both sides by $xy$: $$y + x \frac{dy}{dx} + xy = 8x xy$$ Step 3: Rearrange and solve for $\frac{dy}{dx}$: $$x \frac{dy}{dx} = 8x^2 y - y - xy = y(8x^2 - 1 - x)$$ $$\frac{dy}{dx} = \frac{y(8x^2 - 1 - x)}{x}$$ Step 4: Higher derivatives require implicit differentiation again. (b) Given $y^2 = \ln(x + y)$. Step 1: Differentiate implicitly: $$2y \frac{dy}{dx} = \frac{1 + \frac{dy}{dx}}{x + y}$$ Step 2: Multiply both sides by $x + y$: $$2y (x + y) \frac{dy}{dx} = 1 + \frac{dy}{dx}$$ Step 3: Rearrange to solve for $\frac{dy}{dx}$. (c) Given $y = (1 + e^{3x})^2$. Step 1: Use chain rule: $$y' = 2(1 + e^{3x}) \cdot 3 e^{3x} = 6 e^{3x} (1 + e^{3x})$$ Step 2: Differentiate again for $y''$ and $y'''$. (d) Given $x \sqrt{y + x^2} = x \sqrt{x + y^2}$. Step 1: Differentiate implicitly and solve for $\frac{dy}{dx}$ at $(3,3)$. --- Consumption function: $$C = \frac{0.8 \sqrt{I} - 0.3 I + 10 \sqrt{I^3}}{2 + 3 \sqrt{I}}$$ Step 1: Rewrite $\sqrt{I^3} = I^{3/2}$. Step 2: Use quotient rule to find $\frac{dC}{dI}$. --- Final answers are given in the detailed steps above for each part.