If \( y=\log(\log x) \), then \( \displaystyle\frac{d^2y}{dx^2} \) is equal to: (Consider \( \log_e x=\log x \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -\displaystyle\frac{(1+\log x)}{(x\log x)^2} \)
By the chain rule:
\( \displaystyle \frac{dy}{dx}=\frac{1}{\log x}\cdot\frac{1}{x} \)
\( \displaystyle =(x\log x)^{-1} \)
Differentiating again, with the product rule for \( x\log x \):
\( \displaystyle \frac{d}{dx}(x\log x)=\log x+x\cdot\frac{1}{x}=1+\log x \)
\( \displaystyle \frac{d^2y}{dx^2}=-(x\log x)^{-2}\cdot(1+\log x) \)
\( \displaystyle =-\frac{(1+\log x)}{(x\log x)^2} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{x\log x} \) — Incorrect. This is \( \frac{dy}{dx} \), the first derivative, not the second.
Option (3) \( \displaystyle\frac{x}{\log x} \) — Incorrect. The second derivative is \( -\frac{(1+\log x)}{(x\log x)^2} \), not \( \frac{x}{\log x} \).
Option (4) \( \displaystyle\frac{(1+\log x)}{x\log x} \) — Incorrect. The minus sign and the square in the denominator from differentiating \( (x\log x)^{-1} \) are both missing.