Question

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

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

Answer 1

Answer: (B), (D)

$y=x^{(\ln x{)}^{\ln (\ln x)}}$
${\lambda }_{ny}={\left({\lambda }_{n}x\right)}^{{\lambda }_n(\lambda \mathrm{mx})}\cdot {\lambda }_{nx}$
${\lambda }_n\left({\lambda }_n\right)={\lambda }_n\left({\lambda }_{n}x\right)\cdot {\lambda }_n\left({\lambda }_{nx}\right)+{\lambda }_n\left({\lambda }_{nx}\right)$
$\frac{1}{\lambda ny}\cdot \frac{1}{y}\frac{dy}{dx}=\frac{2\lambda n(\lambda nx)}{\lambda nx}\cdot \frac{1}{x}+\frac{1}{x\lambda nx}$
$=\frac{2\lambda n(\lambda nx)+1}{x\lambda nx}$
$\therefore \frac{dy}{dx}=\frac{y}{x}\cdot \frac{\lambda ny}{\lambda nx}(2\lambda n(\lambda nx)+1)\Rightarrow D$
Substituting the value of $\ln y$ from (1)
$\frac{dy}{dx}=\frac{y}{x}(\lambda nx{)}^{\lambda n(\lambda nx)}(2\lambda n(\lambda nx)+1)\Rightarrow B$