Question

Find the second order derivatives of the function
$log(logx)$

Answer 1

The correct answer is $\frac{-(1+logx)}{(xlogx)^2}$
Let $y=log(logx)$
Then,
$\frac{dy}{dx}=\frac{d}{dx}log(logx)=\frac{1}{logx}.\frac{d}{dx}(logx)=\frac{1}{xlogx}=(xlogx)^{-1}$
$∴\frac{d^2y}{dx^2}=\frac{d}{dx}[(xlogx)^{-1}]$
$=-1(xlogx)^{-2}.\frac{d}{dx}(xlogx)$
$=(\frac{-1}{(xlogx)^2}).[logx.\frac{d}{dx}(x)+x.\frac{d}{dx}(logx)]$
$=\frac{-1}{(xlogx)^2}.[logx.1+x.\frac{1}{x}]=\frac{-(1+logx)}{(xlogx)^2}$