Question

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

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

Answer 1

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

$y=x^{(\ln x{)}^{\ln \left(\ln x\right)}}$
Taking $\ln$ both sides, we get
$\Rightarrow \ln y=\left(\ln x\right)(\ln x{)}^{\ln \left(\ln x\right)}\cdots (1)$
Again, taking $\ln$ of both sides, we get
$\Rightarrow \ln \left(\ln y\right)=\ln \left(\ln x\right)+\ln \left(\ln x\right)\ln \left(\ln x\right)$
$\Rightarrow \ln \left(\ln y\right)=\ln \left(\ln x\right)+\left(\ln \right(\ln x){)}^2$

Differentiating w.r.t. $x$ we get
$\frac{1}{\ln y}\cdot \frac{1}{y}\frac{dy}{dx}=\frac{1}{x\ln x}+\frac{2\ln \left(\ln x\right)}{\ln x}\frac{1}{x}$
$\Rightarrow \frac{1}{\ln y}\cdot \frac{1}{y}\frac{dy}{dx}=\frac{2\ln \left(\ln x\right)+1}{x\ln x}$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}\cdot \frac{\ln y}{\ln x}\left(2\ln \right(\ln x)+1)$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}\left(\ln x{)}^{\ln \left(\ln x\right)}\right(2\ln \left(\ln x\right)+1)\text{(Using (1))}$