Question

If $y = {x^{{x^{{x^{x...\infty }}}}}}$ then $\dfrac{{dy}}{{dx}} =$?

$$\left( A \right)\quad \dfrac{y}{{x\left( {1 - \log x} \right)}} \\\ \left( B \right)\quad \dfrac{{{y^2}}}{{x\left( {1 - \log x} \right)}} \\\ \left( C \right)\quad \dfrac{{{y^2}}}{{x\left( {1 - y\log x} \right)}} \\\ \left( D \right)\quad none\;of\;these \\\$$

Answer 1

We will first convert the given equation in a familiar one (exponential, linear etc). Here we will convert the equation in exponential form. After that take the log on both sides and differentiate.

Complete step-by-step answer:
We are given that $y = {x^{{x^{{x^{x...\infty }}}}}}$. If we look closely, we can observe that,

$$\Rightarrow y = {x^{{x^{{x^{x...\infty }}}}}} \\\ \Rightarrow y = {\left( x \right)^{{x^{{x^{x...\infty }}}}}} \\\ \Rightarrow y = {\left( x \right)^y} \\\ \Rightarrow y = {x^y} \\\$$

As we know, when we have to find the $\dfrac{{dy}}{{dx}}$ of exponential functions, we first take the $\log$ on both sides of equations and proceed further.
$\Rightarrow y = {x^y}$
We know that $\log \left( {{m^n}} \right) = n\log \left( m \right)$.
Taking $\log$ both sides, we get
$\Rightarrow \log \left( y \right) = \log \left( {{x^y}} \right) \\\ \Rightarrow \log \left( y \right) = y\log \left( x \right) \\\$
Now, we will differentiate the equation on both sides.
$\Rightarrow \dfrac{{d\left( {\log \left( y \right)} \right)}}{{dx}}\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {y\log \left( x \right)} \right)}}{{dx}}$
As we know, $\dfrac{{d\left( {u.v} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
Therefore,
$\Rightarrow \dfrac{{d\left( {\log \left( y \right)} \right)}}{{dx}}\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {y\log \left( x \right)} \right)}}{{dx}} \\\ \Rightarrow \left( {\dfrac{1}{y}} \right)\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dx}}\left( {\log \left( x \right)} \right) + y\left( {\dfrac{1}{x}} \right) \\\$
On simplifying this, we will get
$\Rightarrow \left( {\dfrac{1}{y}} \right)\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dx}}\left( {\log \left( x \right)} \right) + y\left( {\dfrac{1}{x}} \right) \\\ \Rightarrow \left( {\dfrac{1}{y} - \log x} \right)\dfrac{{dy}}{{dx}} = \dfrac{y}{x} \\\ \Rightarrow \left( {\dfrac{{1 - y\log x}}{y}} \right)\dfrac{{dy}}{{dx}} = \dfrac{y}{x} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \left( {\dfrac{y}{x}} \right)\left( {\dfrac{y}{{1 - y\log x}}} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{x\left( {1 - y\log x} \right)}} \\\$
Hence the answer is $\left( C \right)\quad \dfrac{{{y^2}}}{{x\left( {1 - y\log x} \right)}}$.
Additional information : Some of the properties and useful formulas are as follows.
$\Rightarrow \log \left( {mn} \right) = \log \left( m \right) + \log \left( n \right) \\\ \Rightarrow \log \left( {{m^n}} \right) = n\log \left( m \right) \\\$

Note: It is necessary to convert the given form of $y = {x^{{x^{{x^{x...\infty }}}}}}$in the form$y = {x^y}$. Otherwise we won’t be able to solve the question properly.