1. Given problem 2.2.1: Find $\frac{dy}{dx}$ for $y = \cos\left(\frac{x - 1}{1 + x^2}\right)$.\n\n2. Use the chain rule: $\frac{dy}{dx} = -\sin\left(\frac{x - 1}{1 + x^2}\right) \cdot \frac{d}{dx}\left(\frac{x - 1}{1 + x^2}\right)$.\n\n3. Differentiate the inner function using quotient rule: If $u = x-1$ and $v = 1 + x^2$, then $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} = \frac{(1 + x^2)(1) - (x-1)(2x)}{(1 + x^2)^2}$.\n\n---\n\n4. Given problem 2.2.2: Find $\frac{dy}{dx}$ for $y = \tan\left[\sqrt[3]{x^2} + \ln(5x^4)\right]$.\n\n5. Let $z = \sqrt[3]{x^2} + \ln(5x^4)$, then $\frac{dy}{dx} = \sec^2(z) \cdot \frac{dz}{dx}$.\n\n6. Differentiate $z$: $\frac{d}{dx}\sqrt[3]{x^2} = \frac{d}{dx} x^{2/3} = \frac{2}{3}x^{-1/3}$ and $\frac{d}{dx}\ln(5x^4) = \frac{1}{5x^4} \cdot 20x^3 = \frac{20x^3}{5x^4} = \frac{4}{x}$.\n\n7. Summing, $\frac{dz}{dx} = \frac{2}{3}x^{-1/3} + \frac{4}{x}$.\n\n---\n\n8. Given problem 2.3: Find $\frac{dy}{dx}$ for $y = x^{\frac{1}{2}}(2 - 3x)^x$ using logarithmic differentiation.\n\n9. Take natural logs: $\ln y = \ln\left(x^{\frac{1}{2}}\right) + \ln\left((2-3x)^x\right) = \frac{1}{2}\ln x + x \ln (2-3x)$.\n\n10. Differentiate implicitly: $\frac{1}{y} \frac{dy}{dx} = \frac{1}{2x} + \ln(2-3x) + x \cdot \frac{-3}{2 - 3x} = \frac{1}{2x} + \ln(2-3x) - \frac{3x}{2 - 3x}$.\n\n11. Multiply both sides by $y$: $\frac{dy}{dx} = y \left( \frac{1}{2x} + \ln(2-3x) - \frac{3x}{2 - 3x} \right) = x^{\frac{1}{2}}(2 - 3x)^x \left( \frac{1}{2x} + \ln(2-3x) - \frac{3x}{2 - 3x} \right)$.