Differentiating the Logarithmic Function

The problem asks us to find the derivative of the function $f(x) = -\log(\log x)$ with respect to $x$. The condition $x > 1$ ensures that $\log x$ is defined and positive, which is necessary for $\log(\log x)$ to be defined.

Understanding the Differentiation Task

We need to compute $\frac{d}{dx}[-\log(\log x)]$. This requires using the chain rule, as the function is a composition of two functions: the outer function is $-\log(u)$ and the inner function is $u = \log x$.

Applying the Chain Rule

The chain rule states that if we have a composite function $y = f(g(x))$, then its derivative is $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

In our case:

• Let $y = -\log(u)$ where $u = \log x$.
• First, find the derivative of the outer function $y = -\log(u)$ with respect to $u$: $$ \frac{dy}{du} = \frac{d}{du}(-\log u) = -\frac{1}{u} $$
• Next, find the derivative of the inner function $u = \log x$ with respect to $x$: $$ \frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x} $$
• Now, apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. $$ \frac{dy}{dx} = \left(-\frac{1}{u}\right) \times \left(\frac{1}{x}\right) $$
• Substitute $u = \log x$ back into the expression: $$ \frac{dy}{dx} = \left(-\frac{1}{\log x}\right) \times \left(\frac{1}{x}\right) $$
• Simplify the expression: $$ \frac{dy}{dx} = -\frac{1}{x \log x} $$

Final Answer

The derivative of {$-\log (\log x)$} with respect to $x$ is {$-\frac{1}{x \log x}$}. This matches the first option provided.