Question

Differentiate \(\log(\log x)\), \(x>1\) with respect to \(x\).

Hint:

Chain rule: d/dx[log(log x)] = (1/log x)·(1/x).

Answer 1

Step 1: Rename the inner function to make the chain rule visible:
Let $t=\log x$, so we need $\dfrac{d}{dx}\log t$.

Step 2: Differentiate with respect to $t$ first, then convert back to $x$:
$\dfrac{d(\log t)}{dt}=\dfrac1t$, and by the chain rule $\dfrac{d(\log t)}{dx}=\dfrac1t\cdot\dfrac{dt}{dx}$.

Step 3: Substitute $\dfrac{dt}{dx}=\dfrac1x$ and $t=\log x$ back in:
$\dfrac{d}{dx}\log(\log x) = \dfrac{1}{\log x}\cdot\dfrac1x = \dfrac{1}{x\log x}$.

Final Answer:
The required derivative is $\dfrac{1}{x\log x}$. \[ \boxed{\dfrac{1}{x\log x}} \]