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# Differentiate ${x^x}$ with respect to x log x.(A) ${x^x}$(B) $\dfrac{1}{x}$(C) ${x^{x + 1}}$(D) None of these  Hint: We can use chain Rule and Product Rule to differentiate both sides. We assume that $u = {x^2}\,\,\,\nu = x\,\,\log \,\,x$. Thereafter, taking log both sides of both values and then we will differentiate the value with respect to x.

Complete step by step solution:
$u = {x^x}\,\,\,\nu = \log \,x$
$u = {x^x}\,$
Taking log on both sides, we will get
$u = {x^x}\, \\ \log u = \log {x^x} \\ \log u = x\log x \\$ $\left( {\because \log \,{a^m} = m\log a} \right)$
Now differentiate both sides, we will get
$\dfrac{d}{{dx}}\,\,\left( {\log \,\,u} \right) = \dfrac{d}{{dx}}\left( {x\log x} \right)$
$\dfrac{1}{u}\,\,\dfrac{{du}}{{dx}} = \dfrac{d}{x}\left( x \right)\,\,.\,\,\log x + x\,\,.\,\,\dfrac{d}{{dx}}\left( {\log x} \right)\,\,\,\left[ {Chain\,\,Rule} \right]$
$\dfrac{{du}}{{dx}} = \,\,u\left[ {\log \,x + \dfrac{x}{x}} \right]$
As, we know that $u = {x^x}$
$\dfrac{{du}}{{dx}} = \,\,{x^x}\left( {\log x + 1} \right)$ $.......(i)$
We will solve $\nu = x\log x$ part
$\nu = x\log x$
$\dfrac{d}{{dx}}\left( \nu \right) = \dfrac{{dx}}{{dx}}\left( {x\left( {\log x} \right)} \right)\,\,\,\left[ {{\text{Product}}\,\,{\text{Rule}}} \right]$
$\dfrac{{d\nu }}{{dx}} = x\dfrac{d}{{dx}}\left( {\log x} \right)\, + \log x\,\,.\,\,\dfrac{d}{{d\left( x \right)}}\left( x \right)$
$= x\dfrac{1}{x} + \log \left( x \right).1$
$\dfrac{{d\nu }}{{dx}} = 1 + \log x$ $........(ii)$
Divide equation $(i)$and $(ii)$ , we have
$\dfrac{{du}}{{d\nu }} = \dfrac{{d\nu /dx}}{{d\nu /dx}} \\ = \dfrac{{{x^2}\left( {\log x + 1} \right)}}{{\left( {\log x + 1} \right)}} \\$
$= {x^x}\,\,Answer$

Option (A) is correct

Note: Chain rule states that the derivative of $f\left[ {g\left( x \right)} \right]$ is${f^1}\left[ {g\left( x \right)} \right]{g^1}\left( x \right)$.
Product rule is also a formula of products used to find the derivatives of products of two or more functions.

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