Question

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

• A. $\left\{{\left(\mathrm{xlogx}\right)}^{\log \mathrm{logx}}\left\{\left(\frac{1}{\mathrm{xlogx}}\right)\left(\mathrm{logx}+\log \mathrm{logx}\right)\right\}+\left(\log \mathrm{logx}\right)\left[\left(\frac{1}{\mathrm{x}}\right)+\left(\frac{1}{\mathrm{xlogx}}\right)\right]\right\}$
• B. $\left\{{\left(\mathrm{xlogx}\right)}^{\log \mathrm{logx}}\left[\log (\log x)\right]\left[\left(\frac{2}{\mathrm{logx}}\right)+\frac{1}{x}\right]\right\}$
• C. ${\left(\mathrm{xlogx}\right)}^{\mathrm{xlogx}}\left[\frac{\left(\log \mathrm{logx}\right)}{\mathrm{x}}\right]\left[\left(\frac{1}{\mathrm{logx}}\right)+1\right]$
• D. $\left[\frac{\mathrm{ylogy}}{\mathrm{xlogx}}\right]\left[2\log \left(\mathrm{logx}\right)+1\right]$

Answer 1

The correct option is D

$\left[\frac{\mathrm{ylogy}}{\mathrm{xlogx}}\right]\left[2\log \left(\mathrm{logx}\right)+1\right]$

Explanation for the correct option:

Step 1: Simplifying the given equation:

Given that,

$\mathrm{y}={\left({\mathrm{x}}^{\mathrm{logx}}\right)}^{\log \log \mathrm{x}}$

By taking $\log$ on both sides,

$\begin{array}{l}\mathrm{logy}=\log \left({\mathrm{x}}^{{\left(\mathrm{logx}\right)}^{\left(\log \left(\mathrm{logx}\right)\right)}}\right)\\ ={\left(\mathrm{logx}\right)}^{\log \left(\mathrm{logx}\right)}.\log x\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{loga}}^{\mathbf{b}}\mathbf{=}\mathbf{bloga}\mathbf{]}\end{array}$

Now, by taking $\log$ again on both sides,

$$\begin{gathered} \log \left(\mathrm{logy}\right)=\log \left[{\left(\mathrm{logx}\right)}^{\log \left(\mathrm{logx}\right)}.\left(\mathrm{logx}\right)\right] \\ \begin{array}{c}=\log \left({\mathrm{logx}}^{\log \left(\mathrm{logx}\right)}\right)+\log \left(\mathrm{logx}\right)\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{log}\mathbf{\left(}\mathbf{ab}\mathbf{\right)}\mathbf{=}\mathbf{loga}\mathbf{+}\mathbf{logb}\mathbf{]}\\ =\log \left(\mathrm{logx}\right).\log \left(\mathrm{logx}\right)+\log \left(\mathrm{logx}\right)\\ ={\left(\log \left(\mathrm{logx}\right)\right)}^2+\left(\log \left(\mathrm{logx}\right)\right)...\left(1\right)\end{array} \end{gathered}$$

Step 2: Find the $\frac{\mathrm{dy}}{\mathrm{dx}}$:

Differentiate equation $\left(1\right)$ both sides with respect to $\mathrm{x}$, we get

$\begin{array}{c}\frac{1}{\mathrm{logy}}.\frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}=2\left(\log \left(\mathrm{logx}\right)\right).\frac{1}{\mathrm{logx}}.\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{logx}}\times \frac{1}{\mathrm{x}}\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{dx}}\left(\mathrm{logx}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{]}\\ \Rightarrow \frac{1}{\mathrm{ylogy}}.\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{xlogx}}\left(2\log \left(\mathrm{logx}\right)+1\right)\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{\mathrm{ylogy}}{\mathrm{xlogx}}\right]\left(2\log \left(\mathrm{logx}\right)+1\right)\end{array}$

Hence, the correct option is D.