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# Differentiate $\dfrac{{x\log x}}{{{e^x}}}$ with respect to $x$

Last updated date: 12th Sep 2024
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Hint: The problem can be solved with the Substitution Method. We have to substitute $\log x = t$. So, we will get $x = {e^t}$. Then put this value in the equation given the question and differentiate it.

Firstly, we will differentiate $\log x$
Substitute t at the place of $\log x$
$\Rightarrow \log x = t$
$\Rightarrow x = {e^t}$
Then, differentiating both sides, we get
$\Rightarrow d\log x = dt$
$\Rightarrow \dfrac{1}{x}dx = dt$
$\Rightarrow \left( {\dfrac{{dt}}{{dx}}} \right) = x$
Putting the above value in $\dfrac{{x\log x}}{{{e^x}}}$, we get
$\Rightarrow \dfrac{{{e^t}t}}{{{e^{{e^t}}}}} = {e^{(t - {e^t})}}t$
Differentiating w.r.t t,
Since the differentiation of ${e^x}$ is ${e^x}$
$\Rightarrow 1.{e^{t - {e^t}}} + {e^{t - {e^t}}}(1 - {e^t})$
Further, we know that $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{1}{x}$
So, put the value of t and multiply the whole equation by $\dfrac{1}{x}$.
$\Rightarrow {\left( e \right)^{t - {e^t}}}(1 + t(1 - {e^t}))\dfrac{1}{x}$
$\Rightarrow \dfrac{{{e^t}}}{{{e^{{e^t}}}}}(1 + t - t{e^t})\dfrac{1}{x}$
Putting the value $t = \log x$ in the above equation, we get
$\Rightarrow \dfrac{1}{{{e^x}}}(1 + \log x - x\log x)$

Therefore $\dfrac{{x\log x}}{{{e^x}}}$ is equal to $\dfrac{1}{{{e^x}}}(1 + \log x - x\log x)$.

The differentiation of ${e^x}\log x$
$\Rightarrow \dfrac{d}{{dx}}({e^x}\log x)$
$\Rightarrow {e^x}\dfrac{d}{{dx}}(\log x) + \log x\dfrac{d}{{dx}}({e^x})$
$\Rightarrow {e^x}\left( {\dfrac{1}{x}} \right) + logx({e^x})$
$\Rightarrow {e^x}(\log x + \dfrac{1}{x})$