\[ \frac{d}{dx} [f(g(x))] = f'(g(x)) g'(x) \] Given Function: \[ -\log(\log x), \quad x > 1 \] Let \(\log x = y\), then differentiating both sides with respect to \(x\): \[ \Rightarrow \frac{dy}{dx} = \frac{1}{x} \quad \text{...(1)} \] Now, given function can be rewritten as: \[ -\log(\log x) = -\log y \] Differentiating both sides with respect to \(x\): \[ \frac{d}{dx} (-\log y) = -\frac{1}{y} \cdot \frac{dy}{dx} \] Substituting from equation (1): \[ = -\frac{1}{\log x} \cdot \frac{1}{x} \] \[ = -\frac{1}{x \log x} \] Thus, the correct derivative is: \[ \frac{d}{dx} [-\log(\log x)] = -\frac{1}{x \log x} \] Hence, option (1) is correct. ✅