1. Given the implicit function $x^2 y^3 = 8$, find $\frac{dy}{dx}$ at $x = -1$. Differentiate both sides implicitly using the product and chain rules: $$2x y^3 + x^2 \cdot 3 y^2 \frac{dy}{dx} = 0$$ Solve for $\frac{dy}{dx}$: $$3 x^2 y^2 \frac{dy}{dx} = -2 x y^3$$ $$\frac{dy}{dx} = \frac{-2 x y^3}{3 x^2 y^2} = \frac{-2 y}{3 x}$$ At $x = -1$, find $y$ from $x^2 y^3 = 8$: $$1^2 y^3 = 8 \implies y^3 = 8 \implies y = 2$$ Evaluate $\frac{dy}{dx}$ at $x = -1$, $y = 2$: $$\frac{dy}{dx} = \frac{-2 \times 2}{3 \times (-1)} = \frac{-4}{-3} = \frac{4}{3}$$ Answer: a) $\frac{4}{3}$ 2. For $3^y = \sin x + \cos x$, find $\frac{dy}{dx}$. Take natural logs or differentiate implicitly: $$3^y = \sin x + \cos x$$ Differentiate both sides: $$3^y \ln 3 \cdot \frac{dy}{dx} = \cos x - \sin x$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{\cos x - \sin x}{3^y \ln 3} = \frac{\cos x - \sin x}{3^y \ln 3}$$ Rewrite in terms of $y$: $$\frac{dy}{dx} = y \cdot \frac{\cos x - \sin x}{\ln 3}$$ Answer: b) $\frac{y}{\ln 3} (\cos x - \sin x)$ 3. Given $x^3 + y^2 -7x + 5y = 8$, find $\frac{dy}{dx}$. Differentiate implicitly: $$3x^2 + 2y \frac{dy}{dx} - 7 + 5 \frac{dy}{dx} = 0$$ Group $\frac{dy}{dx}$ terms: $$2y \frac{dy}{dx} + 5 \frac{dy}{dx} = 7 - 3x^2$$ $$\frac{dy}{dx} (2y + 5) = 7 - 3x^2$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{7 - 3x^2}{2y + 5}$$ Answer: c) $\frac{7 - 3x^2}{2y + 5}$ 4. Given $y = n^3 -1$ and $z = 1 - n^2$, find $\frac{dy}{dz}$ where $n \neq 0$. First find $\frac{dy}{dn}$ and $\frac{dz}{dn}$: $$\frac{dy}{dn} = 3n^2$$ $$\frac{dz}{dn} = -2n$$ Then: $$\frac{dy}{dz} = \frac{\frac{dy}{dn}}{\frac{dz}{dn}} = \frac{3n^2}{-2n} = -\frac{3}{2} n$$ Answer: b) $-\frac{3}{2} n$ 5. Given parametric equations $x = \sqrt{3t - 2}$ and $y = \sqrt{4t + 1}$, find $\frac{dy}{dx}$ at $t=2$. Differentiate: $$\frac{dx}{dt} = \frac{1}{2}(3t - 2)^{-1/2} \cdot 3 = \frac{3}{2 \sqrt{3t - 2}}$$ $$\frac{dy}{dt} = \frac{1}{2}(4t + 1)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4t + 1}}$$ So, $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2 / \sqrt{4t + 1}}{3 / (2 \sqrt{3t - 2})} = \frac{2}{\sqrt{4t + 1}} \cdot \frac{2 \sqrt{3t - 2}}{3} = \frac{4 \sqrt{3t - 2}}{3 \sqrt{4t + 1}}$$ Evaluate at $t=2$: $$x:\sqrt{3(2) - 2} = \sqrt{6 - 2} = \sqrt{4} = 2$$ $$y:\sqrt{4(2) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3$$ So, $$\frac{dy}{dx} = \frac{4 \times 2}{3 \times 3} = \frac{8}{9}$$ Answer: d) $\frac{8}{9}$