Question

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

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

Answer 1

Answer: (B), (D)

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

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

Taking log on both sides,

$lny=ln\left(lnx\right)(lnx{)}^2$$\frac{y^{\prime }}{y}=ln\left(lnx\right)(2lnx.\frac{1}{x})+\left(lnx{)}^2\right(\frac{1}{xlnx})$

$\Rightarrow \frac{y^{\prime }}{y}=ln\left(lnx\right)(2lnx.\frac{1}{x})+\left(lnx\right)\left(\frac{1}{x}\right)$

$\Rightarrow y^{\prime }=y\left(lnx\right)\left(\frac{1}{x}\right)\left(2ln\right(lnx)+1)$

$\Rightarrow y^{\prime }=\frac{y}{x}\left(ln{x}^{lnlnx}\right)\left(2ln\right(lnx)+1)$

We know, $\Rightarrow lny=ln{x}^{lnlnx}.lnx$

So,$y^{\prime }=\frac{ylny}{xlnx}\left(2ln\right(lnx)+1)$