(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 \(x = e^{\frac{2}{t}} \to e^0 = 1\).