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Question

Answers

(A) \[{x^x}\]

(B) \[\dfrac{1}{x}\]

(C) \[{x^{x + 1}}\]

(D) None of these

Answer
Verified

\[u = {x^x}\,\,\,\nu = \log \,x\]

\[u = {x^x}\,\]

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)}} \\

\]

\[ = {x^x}\,\,Answer\]

Product rule is also a formula of products used to find the derivatives of products of two or more functions.

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