Question

If $y = \log (\log \, x)$ then $\frac{d^2 y}{dx^2}$ 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{d y}{d x} =\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) $
$=\frac{-(1+\log x)}{(x \log x)^{2}}$