Differentiate {-log (log x), x > 1} with respect to x

Concept

Here log denotes the natural logarithm (base e), so d/dx(log x) = 1/x. For a composite function f(g(x)), the chain rule gives d/dx f(g(x)) = f'(g(x)) · g'(x). A constant multiplier (here the leading minus sign) is carried through differentiation unchanged.

Application

1. Identify the composite: the outer function is -log(u) with inner u = log x.

2. Differentiate the outer: d/du[-log u] = -1/u.

3. Differentiate the inner: du/dx = d/dx(log x) = 1/x.

4. Multiply by the chain rule: (-1/u)·(1/x) = (-1/log x)·(1/x).

5. Combine into one fraction: = -1 / (x log x).

Cross-check

Sanity check the sign and magnitude: for x > 1 both x > 0 and log x > 0, so x log x > 0 and the result -1/(x log x) is negative. That matches the original function -log(log x), which is decreasing for x > 1 (as x grows, log x grows, log(log x) grows, so -log(log x) falls). A negative derivative is exactly what a decreasing function requires.

Result

d/dx[-log(log x)] = -1 / (x log x).