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_{n y}=\left(\lambda_n x\right)^{\lambda_n(\lambda \operatorname{mx})} \cdot \lambda_{n x} $
$\lambda_n\left(\lambda_n\right)=\lambda_n\left(\lambda_n x\right) \cdot \lambda_n\left(\lambda_{n x}\right)+\lambda_n\left(\lambda_{n x}\right)$
$\frac{1}{\lambda n y} \cdot \frac{1}{y} \frac{d y}{d x}=\frac{2 \lambda n(\lambda n x)}{\lambda n x} \cdot \frac{1}{x}+\frac{1}{x \lambda n x} $
$=\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)
$\frac{d y}{d x}=\frac{y}{x}(\lambda n x)^{\lambda n(\lambda n x)}(2 \lambda n(\lambda n x)+1) \Rightarrow B$