Question

If $y=\log (\log x)$ then $\frac{d^{2}y}{d{x}^2}$ is equal to

• A. $\frac{-\left(1+\log x\right)}{{\left(x\log x\right)}^2}$
• B. $\frac{-\left(1+\log x^2\right)}{x^2\log x}$
• C. $\frac{\left(1+\log x\right)}{{\left(x\log x\right)}^2}$
• D. $\frac{\left(1+\log x^2\right)}{x^2\log x}$

Answer 1

Answer: (A)

We have,
$y=\log (\log x)$
$\frac{dy}{dx}=\frac{1}{\log x}\cdot \frac{1}{x}$
$=\frac{1}{x\log x}$
$=(x\log x{)}^{-1}$
Again differentiating w.r.t. $x$, we get
$\frac{d^{2}y}{d{x}^2}=-(x\log x{)}^{-1-1}\left[1\cdot \log x+x\cdot \frac{1}{x}\right]$
$=-(x\log x{)}^{-2}(\log x+1)$
$=\frac{-(1+\log x)}{(x\log x{)}^2}$