1. **Problem 29:** Find the derivative of \(y = (x^2 + 2)(x^4 + 4)^4\) using logarithmic differentiation. 2. Take the natural logarithm of both sides: $$\ln y = \ln\left((x^2 + 2)(x^4 + 4)^4\right)$$ 3. Use logarithm properties to separate terms: $$\ln y = \ln(x^2 + 2) + \ln\left((x^4 + 4)^4\right) = \ln(x^2 + 2) + 4\ln(x^4 + 4)$$ 4. Differentiate both sides with respect to \(x\) using implicit differentiation: $$\frac{1}{y} \frac{dy}{dx} = \frac{2x}{x^2 + 2} + 4 \cdot \frac{4x^3}{x^4 + 4} = \frac{2x}{x^2 + 2} + \frac{16x^3}{x^4 + 4}$$ 5. Multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = y \left( \frac{2x}{x^2 + 2} + \frac{16x^3}{x^4 + 4} \right)$$ 6. Substitute back \(y = (x^2 + 2)(x^4 + 4)^4\): $$\frac{dy}{dx} = (x^2 + 2)(x^4 + 4)^4 \left( \frac{2x}{x^2 + 2} + \frac{16x^3}{x^4 + 4} \right)$$ 7. Simplify the expression inside the parentheses: $$= (x^2 + 2)(x^4 + 4)^4 \left( \frac{2x(x^4 + 4) + 16x^3(x^2 + 2)}{(x^2 + 2)(x^4 + 4)} \right)$$ 8. Expand numerator: $$2x(x^4 + 4) = 2x^5 + 8x$$ $$16x^3(x^2 + 2) = 16x^5 + 32x^3$$ 9. Sum numerator terms: $$2x^5 + 8x + 16x^5 + 32x^3 = 18x^5 + 32x^3 + 8x$$ 10. Final derivative: $$\frac{dy}{dx} = (x^2 + 2)(x^4 + 4)^4 \cdot \frac{18x^5 + 32x^3 + 8x}{(x^2 + 2)(x^4 + 4)} = (x^4 + 4)^3 (18x^5 + 32x^3 + 8x)$$ --- 11. **Problem 30:** Find the derivative of \(y = \frac{e^{-x} \cos^2 x}{x^2 + x + 1}\) using logarithmic differentiation. 12. Take the natural logarithm of both sides: $$\ln y = \ln\left( \frac{e^{-x} \cos^2 x}{x^2 + x + 1} \right) = \ln(e^{-x} \cos^2 x) - \ln(x^2 + x + 1)$$ 13. Use logarithm properties: $$\ln y = \ln e^{-x} + \ln \cos^2 x - \ln(x^2 + x + 1) = -x + 2 \ln \cos x - \ln(x^2 + x + 1)$$ 14. Differentiate both sides with respect to \(x\): $$\frac{1}{y} \frac{dy}{dx} = -1 + 2 \cdot \frac{-\sin x}{\cos x} - \frac{2x + 1}{x^2 + x + 1} = -1 - 2 \tan x - \frac{2x + 1}{x^2 + x + 1}$$ 15. Multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = y \left(-1 - 2 \tan x - \frac{2x + 1}{x^2 + x + 1} \right)$$ 16. Substitute back \(y = \frac{e^{-x} \cos^2 x}{x^2 + x + 1}\): $$\frac{dy}{dx} = \frac{e^{-x} \cos^2 x}{x^2 + x + 1} \left(-1 - 2 \tan x - \frac{2x + 1}{x^2 + x + 1} \right)$$