(a) To find \(\frac{d}{d\theta}(\cot^2 \theta)\), use the chain rule:

\(\frac{d}{d\theta}(\cot^2 \theta) = 2 \cot \theta \cdot \frac{d}{d\theta}(\cot \theta)\).

Given \(\frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta\), substitute:

\(\frac{d}{d\theta}(\cot^2 \theta) = 2 \cot \theta (-\csc^2 \theta) = -\frac{2 \cot \theta}{\sin^2 \theta}\).

(b) Start with the differential equation:

\(x \sin^2 \theta \frac{dx}{d\theta} = \tan^2 \theta - 2 \cot \theta\).

Separate variables:

\(\int x \, dx = \int \frac{\tan^2 \theta - 2 \cot \theta}{\sin^2 \theta} \, d\theta\).

Integrate both sides:

\(\frac{1}{2}x^2 = \int \frac{\tan^2 \theta}{\sin^2 \theta} \, d\theta - \int \frac{2 \cot \theta}{\sin^2 \theta} \, d\theta\).

\(\frac{1}{2}x^2 = \tan \theta + \cot^2 \theta + C\).

Use the initial condition \(x = 2\) when \(\theta = \frac{1}{4}\pi\):

\(\frac{1}{2}(2)^2 = \tan \frac{1}{4}\pi + \cot^2 \frac{1}{4}\pi + C\).

\(2 = 1 + 1 + C \Rightarrow C = 0\).

Thus, \(\frac{1}{2}x^2 = \tan \theta + \cot^2 \theta\).

Substitute \(\theta = \frac{1}{6}\pi\):

\(\frac{1}{2}x^2 = \tan \frac{1}{6}\pi + \cot^2 \frac{1}{6}\pi\).

Calculate \(x\):

\(x = \sqrt{2(\tan \frac{1}{6}\pi + \cot^2 \frac{1}{6}\pi)} \approx 2.67\).

(a) Let \(y = \ln(\ln x)\). Then \(\frac{dy}{dx} = \frac{d}{dx}(\ln(\ln x))\).

Using the chain rule, \(\frac{dy}{dx} = \frac{1}{\ln x} \cdot \frac{d}{dx}(\ln x) = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}\).

(b) The differential equation is \(x \ln x + t \frac{dx}{dt} = 0\).

Rearrange to get \(t \frac{dx}{dt} = -x \ln x\).

Separate variables: \(\frac{dx}{x \ln x} = -\frac{dt}{t}\).

Integrate both sides: \(\int \frac{1}{x \ln x} \, dx = -\int \frac{1}{t} \, dt\).

The left side integrates to \(\ln(\ln x)\) and the right side to \(-\ln t + C\).

Thus, \(\ln(\ln x) = -\ln t + C\).

Using the initial condition \(x = e\) when \(t = 2\), we find \(\ln(\ln e) = -\ln 2 + C\).

Since \(\ln e = 1\), \(C = \ln 2 + 1\).

Substitute back: \(\ln(\ln x) = -\ln t + \ln 2 + 1\).

Exponentiate to solve for \(x\): \(\ln x = e^{-\ln t + \ln 2 + 1}\).

\(x = e^{\frac{2}{t}}\).

(c) As \(t \to \infty\), \(\frac{2}{t} \to 0\), so \(x = e^{\frac{2}{t}} \to e^0 = 1\).

(a) Separate variables:

\(\frac{dy}{y} = \frac{\sin x}{1 - \cos x} dx\)

Integrate both sides:

\(\int \frac{dy}{y} = \int \frac{\sin x}{1 - \cos x} dx\)

\(\ln y = -\ln(1 - \cos x) + C\)

\(\ln y = \ln(1 - \cos x)^{-1} + C\)

\(y = e^C (1 - \cos x)^{-1}\)

Use initial condition \(y = 4\) when \(x = \pi\):

\(4 = e^C (1 - \cos \pi)^{-1}\)

\(4 = e^C (1 - (-1))^{-1}\)

\(4 = e^C \cdot \frac{1}{2}\)

\(e^C = 8\)

Thus, \(y = 8(1 - \cos x)^{-1}\)

\(y = 2(1 - \cos x)\)

(b) The graph of \(y = 2(1 - \cos x)\) is a sinusoidal curve with a maximum at \(x = \pi\) for \(0 < x < 2\pi\).

(a) To solve the differential equation \(e^{3t} \frac{dx}{dt} = \cos^2 2x\), we first separate variables:

\(\int \sec^2 2x \, dx = \int e^{-3t} \, dt\).

Integrating both sides, we get:

\(\frac{1}{2} \tan 2x = -\frac{1}{3} e^{-3t} + C\).

Using the initial condition \(x = 0\) when \(t = 0\), we find \(C = \frac{1}{3}\).

Thus, \(\frac{1}{2} \tan 2x = -\frac{1}{3} e^{-3t} + \frac{1}{3}\).

Solving for \(x\), we obtain:

\(x = \frac{1}{2} \arctan \left( \frac{2}{3} (1 - e^{-3t}) \right)\).

(b) As \(t \to \infty\), \(e^{-3t} \to 0\), so \(x \to \frac{1}{2} \arctan \left( \frac{2}{3} \right)\).