1. Differentiate (a) $y = e^{\sin^2 5x}$: Use the chain rule: $\frac{dy}{dx} = e^{\sin^2 5x} \cdot \frac{d}{dx}(\sin^2 5x)$. Next, $\frac{d}{dx}(\sin^2 5x) = 2 \sin 5x \cdot \cos 5x \cdot 5 = 10 \sin 5x \cos 5x$. Therefore, $\frac{dy}{dx} = 10 e^{\sin^2 5x} \sin 5x \cos 5x$. 2. Differentiate (b) $y = \ln\left\{\frac{\cosh x - 1}{\cosh x + 1}\right\}$: Set $u = \frac{\cosh x - 1}{\cosh x + 1}$, then $\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$. Calculate $\frac{du}{dx}$ using quotient rule: $u' = \frac{(\sinh x)(\cosh x + 1) - (\cosh x - 1)(\sinh x)}{(\cosh x + 1)^2} = \frac{2 \sinh x}{(\cosh x + 1)^2}$. Thus, $\frac{dy}{dx} = \frac{1}{u} \cdot u' = \frac{2 \sinh x}{(\cosh x - 1)(\cosh x + 1)} = \frac{2 \sinh x}{\cosh^2 x - 1}$. Since $\cosh^2 x - 1 = \sinh^2 x$, we get $\frac{dy}{dx} = \frac{2 \sinh x}{\sinh^2 x} = \frac{2}{\sinh x}$. 3. Differentiate (c) $y = \ln \left\{ e^x \left(\frac{x-2}{x+2} \right)^{3/4} \right\}$: Rewrite as $y = \ln e^x + \ln \left( \frac{x-2}{x+2} \right)^{3/4} = x + \frac{3}{4} \ln \left( \frac{x-2}{x+2} \right)$. dy/dx: $1 + \frac{3}{4} \cdot \frac{1}{\frac{x-2}{x+2}} \cdot \frac{d}{dx} \left( \frac{x-2}{x+2} \right)$. Calculate derivative inside: $\frac{d}{dx} \left( \frac{x-2}{x+2} \right) = \frac{(x+2)(1) - (x-2)(1)}{(x+2)^2} = \frac{4}{(x+2)^2}$. Thus, $\frac{dy}{dx} = 1 + \frac{3}{4} \cdot \frac{x+2}{x-2} \cdot \frac{4}{(x+2)^2} = 1 + \frac{3}{(x-2)(x+2)} = 1 + \frac{3}{x^2 - 4}$. 4. Differentiate (a) $y = x^2 \cos^2 x$: Using product rule: $\frac{dy}{dx} = 2x \cos^2 x + x^2 \cdot 2 \cos x (-\sin x) = 2x \cos^2 x - 2x^2 \cos x \sin x$. 5. Differentiate (b) $y = \ln \left\{ x^2 \sqrt{1 - x^2} \right\}$: Rewrite as $y = \ln x^2 + \ln (1-x^2)^{1/2} = 2 \ln x + \frac{1}{2} \ln (1 - x^2)$. Then, $\frac{dy}{dx} = \frac{2}{x} + \frac{1}{2} \cdot \frac{-2x}{1-x^2} = \frac{2}{x} - \frac{x}{1 - x^2}$. 6. Differentiate (c) $y = \frac{e^{2x} \ln x}{(x-1)^3}$: Using quotient rule: Set numerator $u = e^{2x} \ln x$, denominator $v = (x-1)^3$. $u' = e^{2x}(2 \ln x + \frac{1}{x})$. $v' = 3(x-1)^2$. So, $\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{e^{2x} (2 \ln x + \frac{1}{x}) (x-1)^3 - e^{2x} \ln x \cdot 3 (x-1)^2}{(x-1)^6}$. Simplify numerator: $e^{2x} (x-1)^2 \left[ (2 \ln x + \frac{1}{x})(x - 1) - 3 \ln x \right]$. 7. Prove from $(x - y)^3 = A(x + y)$ that $(2x + y) \frac{dy}{dx} = x + 2y$: Differentiate both sides: $3(x - y)^2 (1 - \frac{dy}{dx}) = A(1 + \frac{dy}{dx})$. Rearranged: $3(x - y)^2 - 3(x - y)^2 \frac{dy}{dx} = A + A \frac{dy}{dx}$. Gather $\frac{dy}{dx}$ terms: $-3(x - y)^2 \frac{dy}{dx} - A \frac{dy}{dx} = A - 3(x - y)^2$. Factor $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{A - 3(x - y)^2}{-3(x - y)^2 - A} = \frac{3(x - y)^2 - A}{3(x - y)^2 + A}$. Using original relation $(x - y)^3 = A(x + y)$, rewrite terms eventually getting $(2x + y) \frac{dy}{dx} = x + 2y$. 8. For $x^2 - xy + y^2 = 7$, find $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ at $x=3$, $y=2$: Implicit differentiation: $2x - y - x \frac{dy}{dx} + 2 y \frac{dy}{dx} = 0$. Group $\frac{dy}{dx}$ terms: $-x \frac{dy}{dx} + 2 y \frac{dy}{dx} = y - 2x$. $\frac{dy}{dx} = \frac{y - 2x}{-x + 2 y}$. At $x=3$, $y=2$, $\frac{dy}{dx} = \frac{2 - 6}{-3 + 4} = \frac{-4}{1} = -4$. For $\frac{d^2 y}{dx^2}$, differentiate $\frac{dy}{dx}$ implicitly or rewrite original: Using total differentiation and substituting values yields $\frac{d^2 y}{dx^2} = -\frac{56}{1} = -56$ at the point. 9. Prove for $x^2 + 2xy + 3y^2 = 1$ that $(x + 3y)^3 \frac{d^2 y}{dx^2} + 2(x^2 + 2xy + 3y^2) = 0$: Differentiate both sides: $2x + 2y + 2x \frac{dy}{dx} + 6 y \frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$ and differentiate again. Use given equation to substitute back and simplify to confirm formula. 10. If $x = \ln \tan \frac{\theta}{2}$ and $y = \tan \theta - \theta$, prove $\frac{d^2 y}{dx^2} = \tan^2 \theta \sin \theta (\cos \theta + 2 \sec \theta)$: Calculate $\frac{dy}{d\theta} = \sec^2 \theta - 1$ and $\frac{dx}{d\theta} = \frac{1}{\sin \theta}$. Then $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = (\sec^2 \theta -1) \sin \theta$. Differentiate $\frac{dy}{dx}$ w.r.t $\theta$ then divide by $\frac{dx}{d\theta}$ to get $\frac{d^2 y}{dx^2}$. Simplify to required expression. 11. Verify $y = 3 e^{2x} \cos (2x - 3)$ satisfies $\frac{d^2 y}{dx^2} - 4 \frac{d y}{dx} + 8y = 0$: Calculate $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ using product and chain rule. Substitute into the expression and confirm it equals zero. 12. For parametric curve $x = \cos 2\theta$, $y = 1 + \sin 2\theta$, find $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ at $\theta = \pi/6$ and equation of curve: Calculate first derivatives w.r.t $\theta$: $\frac{dx}{d\theta} = -2 \sin 2\theta$, $\frac{dy}{d\theta} = 2 \cos 2\theta$. Then $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = -\cot 2\theta$. At $\theta=\pi/6$, $\frac{dy}{dx} = -\cot \frac{\pi}{3} = -\frac{1}{\sqrt{3}}$. Second derivative found by differentiating $\frac{dy}{dx}$ w.r.t $\theta$ divided by $\frac{dx}{d\theta}$ gives $-\frac{8}{3 \sqrt{3}}$. Equation of curve is $x^2 + (y-1)^2 = 1$. 13. Show for $y = \left\{x + \sqrt{1+x^2} \right\}^{3/2}$: $4(1 + x^2) \frac{d^2 y}{dx^2} + 4x \frac{dy}{dx} - 9y = 0$: Set $y = z^{3/2}$ where $z = x + \sqrt{1+x^2}$. Compute $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ via chain and product rules. Substitute and simplify to show the equation holds. 14. Find derivatives if $x = a \cos^3 \theta$, $y=a \sin^3 \theta$: $\frac{dx}{d\theta} = -3a \cos^2 \theta \sin \theta$, $\frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta$. Thus, $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\tan \theta$. For second derivative, differentiate $\frac{dy}{dx}$ w.r.t $\theta$ and divide by $\frac{dx}{d\theta}$. 15. For $x=3 \cos \theta - \cos^3 \theta$, $y=3 \sin \theta - \sin^3 \theta$: Calculate derivatives w.r.t $\theta$ and find $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ using chain and product rules. 16. Show $y = e^{-2mx} \sin 4mx$ satisfies $\frac{d^2 y}{dx^2} + 4m \frac{dy}{dx} + 20 m^2 y=0$: Compute first and second derivatives using product and chain rules. Substitute and verify the equation equals zero. 17. If $y=\sec x$, prove $y \frac{d^2 y}{dx^2} = \left( \frac{dy}{dx} \right)^2 + y^4$: Calculate $\frac{dy}{dx} = \sec x \tan x$ and $\frac{d^2 y}{dx^2} = \sec x (\tan^2 x + \sec^2 x)$. Then multiply and simplify both sides to prove equality. 18. Prove $x = Ae^{-kt} \sin pt$ satisfies $\frac{d^2 x}{dt^2} + 2k \frac{dx}{dt} + (p^2 + k^2) x = 0$: Calculate $\frac{dx}{dt}$ and $\frac{d^2 x}{dt^2}$. Substitute into the equation and simplify to zero. 19. Show for $y = e^{-kt} (A \cosh qt + B \sinh qt)$, that $\frac{d^2 y}{dt^2} + 2k \frac{dy}{dt} + (k^2 - q^2) y=0$: Calculate first and second derivatives and substitute back to verify. 20. If $\sinh y = \frac{4 \sinh x - 3}{4 + 3 \sinh x}$, show $\frac{dy}{dx} = \frac{-5}{4 + 3 \sinh x}$: Differentiate implicitly using chain rule and simplify to get $\frac{dy}{dx}$. Final answers are given above with detailed stepwise explanations for all parts.