Question

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

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

Answer 1

Answer: (B), (D)

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

$y=x^{{\left(\log x\right)}^{\log \left(\log x\right)}}$

Taking log of both sides, we get

$\therefore \log y=\left(\log x\right){\left(\log x\right)}^{\log \left(\log x\right)}$ ... (1)

Taking log of both sides, we get

$\log \left(\log y\right)=\log \left(\log x\right)+\log \left(\log x\right)\log \left(\log x\right)$

Differentiating w.r.t. $x$, we get

$\frac{1}{\log y}.\frac{1}{y}\frac{dy}{dx}=\frac{1}{x\log x}+\frac{2\log \left(\log x\right)}{\log x}\frac{1}{x}$

$=\frac{2\log \left(\log x\right)+1}{x\log x}$

or $\frac{dy}{dx}=\frac{y}{x}.\frac{\log y}{\log x}[2\log \left(\log x\right)+1]$

Substituting the value of $y$ from (1), we get

$\frac{dy}{dx}=\frac{y}{x}{\left(\log x\right)}^{\log \left(\log x\right)}(2\log \left(\log x\right)+1)$