Question

Differentiate \[{x^x}\] with respect to x log x.
(A) \[{x^x}\]
(B) \[\dfrac{1}{x}\]
(C) \[{x^{x + 1}}\]
(D) None of these

Answer 1

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.