Question

If $y = {x^{{x^{x \cdots \infty }}}}$, then find $\dfrac{{dy}}{{dx}}$.
A. $y{x^{y - 1}}$
B. $\dfrac{{{y^2}}}{{x\left( {1 - y\log x} \right)}}$
C. $\dfrac{{{y^2}}}{{x\left( {1 + y\log x} \right)}}$
D. None of these

Answer 1

Hint: We will replace the power of $x$ by $y$. Then take the logarithm function on both sides of the equation. Then differentiate the equation with respect to $x$, to calculate $\dfrac{{dy}}{{dx}}$.

Formula Used:
Logarithm rules:
$\log {a^m} = m\log a$
Derivative formula:
Logarithm rule: $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
Product rule: $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$

Complete step by step solution:
Given equation is $y = {x^{{x^{x \cdots \infty }}}}$.
Since the power of $x$ repeats infinite times.
Replace the power of $x$ by $y$.
$ \Rightarrow y = {x^y}$
Take the logarithm function on both sides of the equation
$ \Rightarrow \log y = \log {x^y}$
Apply the formula of power of logarithm function
$ \Rightarrow \log y = y\log x$
Differentiate both sides of the equation with respect to $x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{d}{{dx}}\left( {y\log x} \right)$
Apply differentiate formulas
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \log x\dfrac{d}{{dx}}\left( y \right) + y\dfrac{d}{{dx}}\left( {\log x} \right)$
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \log x\dfrac{{dy}}{{dx}} + \dfrac{y}{x}$
Combine like terms
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} - \log x\dfrac{{dy}}{{dx}} = \dfrac{y}{x}$
Take common $\dfrac{{dy}}{{dx}}$ from left side expression
$ \Rightarrow \left( {\dfrac{1}{y} - \log x} \right)\dfrac{{dy}}{{dx}} = \dfrac{y}{x}$
Divided both sides by $\left( {\dfrac{1}{y} - \log x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{{x\left( {\dfrac{1}{y} - \log x} \right)}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{{x\left( {\dfrac{{1 - y\log x}}{y}} \right)}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{x\left( {1 - y\log x} \right)}}$

Option ‘B’ is correct

Note: This type of question belongs to the category of function to the function. To solve such a type of question, you need to take the logarithmic function on both sides of the equation and then differentiate both sides with respect to the x. By combining like terms we get the value of $\dfrac{{dy}}{{dx}}$.