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

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.