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# $\dfrac{d}{{dx}}\left( {{x^{\ln x}}} \right)$ is equal to(1) $2{x^{\ln x - 1}}\ln x$(2) ${x^{\ln x - 1}}$(3) $\dfrac{2}{3}\left( {\ln x} \right)$(4) ${x^{\ln x - 1}}.\ln x$

Last updated date: 18th Jun 2024
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Answer
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Hint:
To solve $\dfrac{d}{{dx}}\left( {{x^{\ln x}}} \right)$, first assume that $y = {x^{\ln x}}$. Perform the natural logarithm on both sides. This will give you $\ln y = {\left( {\ln x} \right)^2}$. Now differentiate this equation by using implicit differentiation on the LHS and chain rule on the RHS. You will obtain $\dfrac{1}{y}\dfrac{{dy}}{{dx}} = 2{x^{\ln x - 1}}\ln x$.

Complete step by step solution:
Given: To calculate $\dfrac{d}{{dx}}\left( {{x^{\ln x}}} \right)$
The 4 options given are:
(1) $2{x^{\ln x - 1}}\ln x$ (2) ${x^{\ln x - 1}}$
(3) $\dfrac{2}{3}\left( {\ln x} \right)$ (4) ${x^{\ln x - 1}}.\ln x$
To calculate $\dfrac{d}{{dx}}\left( {{x^{\ln x}}} \right)$
First assume that $y = {x^{\ln x}}$ … (i)
Then we need to evaluate $\dfrac{{dy}}{{dx}}$
Now apply the natural logarithmic function to both sides of the equation (i) to get.
$\ln y = \ln \left( {{x^{\ln x}}} \right)$
$\Rightarrow \ln y = \ln x.\ln x$ $\left[ {\because \ln {a^b} = b\ln a} \right]$
$\Rightarrow \ln y = {\left( {\ln x} \right)^2}$ … (ii)
Now differentiate equation (ii). Apply implicit differentiation on the LHS and chain rule on the RHS to get,
$\dfrac{1}{y}.\dfrac{{dy}}{{dx}} = 2\left( {\ln x} \right)\left( {\dfrac{1}{x}} \right)$ … (iii)
$\Rightarrow \dfrac{1}{{{x^{\ln x}}}}.\dfrac{{dy}}{{dx}} = 2\left( {\ln x} \right){x^{ - 1}} \\ \Rightarrow \dfrac{{dy}}{{dx}} = \left( {{x^{\ln x}}} \right)2\left( {\ln x} \right){x^{ - 1}} \\ \Rightarrow \dfrac{{dy}}{{dx}} = 2{x^{\ln \left( x \right) - 1}}.\ln x \\$

The correct answer for $\dfrac{d}{{dx}}{x^{\ln x}}$ is $2{x^{\ln \left( x \right) - 1}}.\ln x$.

Note:
To solve this kind of question you must substitute a variable as the expression to be differentiated. Here we have assumed$y = {x^{\ln x}}$.
You must know how to handle the formulas of natural logarithms$\left( {\ln } \right)$. You must know to differentiate $\dfrac{d}{{dx}}\ln x$ to efficiently perform chain rule. Here, we have directly performed the chain rule in one step but you may break down each step of the chain reaction and perform it one by one.
For e.g. $\dfrac{d}{{dx}}{\left( {\ln x} \right)^2}$
Assume $u = \ln x$
and $f = {u^2}$
$\dfrac{d}{{dx}}{\left( {\ln x} \right)^2} = \dfrac{{df}}{{du}}.\dfrac{{du}}{{dx}} \\ = \dfrac{d}{{du}}{u^2}.\dfrac{d}{{dx}}.\ln x \\ = 2u.\dfrac{1}{x} \\ = 2\ln x.\dfrac{1}{x} \\$