Question

$x^{m{x}^{m{x}^{m{x}^{......}}}}=y^{n{y}^{n{y}^{n{y}^{......}}}}$, then $\frac{dy}{dx}$is equal to:

• A. $\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.$
• B. $\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.$
• C. $\raisebox{1ex}{$my$}\!\left/ \!\raisebox{-1ex}{$nx$}\right.$
• D. $\raisebox{1ex}{$ny$}\!\left/ \!\raisebox{-1ex}{$mx$}\right.$

Answer 1

The correct option is C

$\raisebox{1ex}{$my$}\!\left/ \!\raisebox{-1ex}{$nx$}\right.$

Explanation for the correct option

Given: $x^{m{x}^{m{x}^{.....\infty }}}=y^{n{y}^{n{y}^{......\infty }}}$

Let $x^{m{x}^{m{x}^{.....\infty }}}=y^{n{y}^{n{y}^{......\infty }}}=t$

Since series is going infinite in power we can write

$x^{mt}=y^{nt}\left[t=x^{m{x}^{x{m}^{....\infty }}}andt=y^{n{y}^{n{y}^{.....\infty }}}\right]$

Take $\log$ on both sides

$\begin{array}{cccc}\Rightarrow & \log \left(x^{mt}\right)& =& \log \left(y^{nt}\right)\end{array}$

$\begin{array}{cccc}\Rightarrow & mt\log \left(x\right)& =& nt\log \left(y\right)\left[\because \log a^b=b\log a\right]\end{array}$

$\begin{array}{cccc}\Rightarrow & m\log \left(x\right)& =& n\log \left(y\right)\end{array}$

Differentiate with respect to $x$

$\begin{array}{cccc}\Rightarrow & m\frac{d}{dx}\left(\log x\right)& =& n\frac{d}{dy}\left(\log y\right)\end{array}$

$\begin{array}{ccccc}\Rightarrow & m\frac{1}{x}& =& n\frac{1}{y}\frac{dy}{dx};& \left[\frac{d}{dx}\left(\log x\right)=\frac{1}{x}\right]\end{array}$

$\begin{array}{cccc}\Rightarrow & \frac{my}{nx}& =& \frac{dy}{dx}\end{array}$

Hence option (c) is the required answer.