Question

If $y=x^{(\log x{)}^{\log \left(\log x\right)}}$, then $\frac{dy}{dx}$ is

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

Answer 1

Answer: (C), (D)

The correct options are
C $\frac{y}{x}\left(\log x{)}^{\log \left(\log x\right)}\right(2\log \left(\log x\right)+1)$
D $\frac{y\log y}{x\log x}\left[2\log \right(\log x)+1]$

$y=x^{(\log x{)}^{\log \left(\log x\right)}}$
Taking $\log$ on both side,
$\log y=\left(\log x\right)(\log x{)}^{\log \left(\log x\right)}\cdots (1)$
Taking $\log$ on both side,
$\log \left(\log y\right)=\log \left(\log x\right)+\log \left(\log x\right)\log \left(\log x\right)$
Differentiating w.r.t. $x$,
$\frac{1}{\log y}\cdot \frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{x\log x}+\frac{2\log \left(\log x\right)}{\log x}\cdot \frac{1}{x}\Rightarrow \frac{1}{\log y}\cdot \frac{1}{y}\cdot \frac{dy}{dx}=\frac{2\log \left(\log x\right)+1}{x\log x}\Rightarrow \frac{dy}{dx}=\frac{y}{x}\cdot \frac{\log y}{\log x}\left(2\log \right(\log x)+1)$
Using equation $\left(1\right)$,
$\frac{dy}{dx}=\frac{y}{x}\left(\log x{)}^{\log \left(\log x\right)}\right(2\log \left(\log x\right)+1)$