Question

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

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

Answer 1

The correct option is D $\frac{y\ln y}{x\ln x}\left(2\ln \left(\ln x\right)+1\right)$

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

$\Rightarrow \ln y={\left(\ln x\right)}^{\ln \left(\ln x\right)}.\ln x$ ...(1)

$\Rightarrow \ln \left(\ln y\right)=\ln \left(\ln x\right).\ln \left(\ln x\right)+\ln \left(\ln x\right)$

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

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

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

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