Question

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

• A. $\frac{-(1+logx)}{x^{2}logx}$
• B. $\frac{(1+logx)}{x^{2}logx}$
• C. $\frac{-(1+logx)}{(xlogx{)}^2}$
• D. $\frac{(1+logx)}{(x^{2}logx{)}^2}$

Answer 1

Answer: (C)

$\frac{-(1+logx)}{(xlogx{)}^2}$

$y=\log \left(\log x\right)$
Again differentiate both sides w.r.t. x
$\frac{dy}{dx}=\frac{1}{\log x}\frac{d}{dx}\left(\log x\right)$
$\frac{dy}{dx}=\frac{1}{\log x}\times \frac{1}{x}\times 1$
$\frac{dy}{dx}=\frac{1}{x\log x}$
Again differentiate both sides w.r.t. x
$\frac{d^{2}y}{d{x}^2}=\frac{x\log x\left(0\right)-1[\log x+x\times \frac{1}{x}]}{{\left(x\log x\right)}^2}$
$\frac{d^{2}y}{d{x}^2}=\frac{-(1+\log x)}{{\left(x\log x\right)}^2}$