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Hint: In this question, the function is the power of the $\log x$ function of x. Therefore, in this case, we can use the chain rule by defining suitable variables and then simplify it to obtain the required answer.

Complete step-by-step answer:

In the question, we have to differentiate ${{\left( \log x \right)}^{x}}$ with respect to $\log x$ i.e. we have to find $\dfrac{d{{\left( \log x \right)}^{x}}}{d\left( \log x \right)}$.

Now, we know that if u and v are two functions of x, then

$\dfrac{du}{dv}=\dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}..............(1.1)$

Therefore, taking $u={{\left( \log x \right)}^{x}}$ and $v=\log x$ in equation (1.1), we obtain

$\dfrac{d{{\left( \log x \right)}^{x}}}{d\left( \log x \right)}=\dfrac{\dfrac{d{{\left( \log x \right)}^{x}}}{dx}}{\dfrac{d\left( \log x \right)}{dx}}=\dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}................(1.2)$

We know that for any numbers a and b

$\log \left( {{a}^{b}} \right)=b\log \left( a \right)..............(1.2)$

Now, we have defined $u$ as $u={{\left( \log x \right)}^{x}}$. Taking logarithm on both sides and using equation (1.2), we obtain

$\log u=\log \left( {{\left( \log x \right)}^{x}} \right)=x\log \left( \log x \right).....(1.3)$

Now, the derivative of log function is given by

$\dfrac{d\log x}{dx}=\dfrac{1}{x}..............(1.4)$

The chain rules is stated as

$\dfrac{d\left( f(g(x)) \right)}{dx}=\dfrac{df(g)}{dg}\times \dfrac{dg(x)}{dx}........(1.5)$

And the derivative of the product of two functions is given by

$\dfrac{d\left( f(x)g(x) \right)}{dx}=g(x)\dfrac{df(x)}{dx}+f(x)\dfrac{dg(x)}{dx}...........(1.6)$

Therefore, differentiating both sides of equation (1.3) and using equations (1.4), (1.5) and (1.6), we get

\[\begin{align}

& \log u=x\log \left( \log x \right) \\

& \Rightarrow \dfrac{d\log u}{dx}=\dfrac{d\left( x\log \left( \log x \right) \right)}{dx} \\

& \Rightarrow \dfrac{d\log u}{du}\dfrac{du}{dx}=\log \left( \log x \right)\dfrac{d\left( x

\right)}{dx}+x\dfrac{d\left( \log \left( \log x \right) \right)}{dx} \\

& \Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\log \left( \log x \right)\times 1+x\times

\dfrac{d\left( \log \left( \log x \right) \right)}{d\log x}\dfrac{d\log x}{dx} \\

& \Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\log \left( \log x \right)+x\times \dfrac{1}{\log x}\times \dfrac{1}{x} \\

& \Rightarrow \dfrac{du}{dx}=u\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)={{\left(

\log x \right)}^{x}}\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right).....(1.7) \\

\end{align}\]

Similarly, in the denominator, we can use equation (1.4) to obtain

$\dfrac{dv}{dx}=\dfrac{d\log x}{dx}=\dfrac{1}{x}..............(1.8)$

Therefore, from equations (1.2), (1.7) and (1.8), we obtain

$\begin{align}

& \dfrac{d{{\left( \log x \right)}^{x}}}{d\left( \log x \right)}=\dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}=\dfrac{{{\left( \log x \right)}^{x}}\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)}{\dfrac{1}{x}} \\

& \Rightarrow \dfrac{d{{\left( \log x \right)}^{x}}}{d\left( \log x \right)}=x{{\left( \log x \right)}^{x}}\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right) \\

\end{align}$

Which is the required answer to this question.

Note: In this case, we should note that we should simply take $\log x$ as a constant and use the derivative of the xth power of a constant as $\dfrac{d\left( {{a}^{x}} \right)}{dx}={{a}^{x}}\log a$ because in this formula a is a constant whereas in the question $\log x$ is not a constant but is a function of x.

Complete step-by-step answer:

In the question, we have to differentiate ${{\left( \log x \right)}^{x}}$ with respect to $\log x$ i.e. we have to find $\dfrac{d{{\left( \log x \right)}^{x}}}{d\left( \log x \right)}$.

Now, we know that if u and v are two functions of x, then

$\dfrac{du}{dv}=\dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}..............(1.1)$

Therefore, taking $u={{\left( \log x \right)}^{x}}$ and $v=\log x$ in equation (1.1), we obtain

$\dfrac{d{{\left( \log x \right)}^{x}}}{d\left( \log x \right)}=\dfrac{\dfrac{d{{\left( \log x \right)}^{x}}}{dx}}{\dfrac{d\left( \log x \right)}{dx}}=\dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}................(1.2)$

We know that for any numbers a and b

$\log \left( {{a}^{b}} \right)=b\log \left( a \right)..............(1.2)$

Now, we have defined $u$ as $u={{\left( \log x \right)}^{x}}$. Taking logarithm on both sides and using equation (1.2), we obtain

$\log u=\log \left( {{\left( \log x \right)}^{x}} \right)=x\log \left( \log x \right).....(1.3)$

Now, the derivative of log function is given by

$\dfrac{d\log x}{dx}=\dfrac{1}{x}..............(1.4)$

The chain rules is stated as

$\dfrac{d\left( f(g(x)) \right)}{dx}=\dfrac{df(g)}{dg}\times \dfrac{dg(x)}{dx}........(1.5)$

And the derivative of the product of two functions is given by

$\dfrac{d\left( f(x)g(x) \right)}{dx}=g(x)\dfrac{df(x)}{dx}+f(x)\dfrac{dg(x)}{dx}...........(1.6)$

Therefore, differentiating both sides of equation (1.3) and using equations (1.4), (1.5) and (1.6), we get

\[\begin{align}

& \log u=x\log \left( \log x \right) \\

& \Rightarrow \dfrac{d\log u}{dx}=\dfrac{d\left( x\log \left( \log x \right) \right)}{dx} \\

& \Rightarrow \dfrac{d\log u}{du}\dfrac{du}{dx}=\log \left( \log x \right)\dfrac{d\left( x

\right)}{dx}+x\dfrac{d\left( \log \left( \log x \right) \right)}{dx} \\

& \Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\log \left( \log x \right)\times 1+x\times

\dfrac{d\left( \log \left( \log x \right) \right)}{d\log x}\dfrac{d\log x}{dx} \\

& \Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\log \left( \log x \right)+x\times \dfrac{1}{\log x}\times \dfrac{1}{x} \\

& \Rightarrow \dfrac{du}{dx}=u\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)={{\left(

\log x \right)}^{x}}\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right).....(1.7) \\

\end{align}\]

Similarly, in the denominator, we can use equation (1.4) to obtain

$\dfrac{dv}{dx}=\dfrac{d\log x}{dx}=\dfrac{1}{x}..............(1.8)$

Therefore, from equations (1.2), (1.7) and (1.8), we obtain

$\begin{align}

& \dfrac{d{{\left( \log x \right)}^{x}}}{d\left( \log x \right)}=\dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}=\dfrac{{{\left( \log x \right)}^{x}}\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)}{\dfrac{1}{x}} \\

& \Rightarrow \dfrac{d{{\left( \log x \right)}^{x}}}{d\left( \log x \right)}=x{{\left( \log x \right)}^{x}}\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right) \\

\end{align}$

Which is the required answer to this question.

Note: In this case, we should note that we should simply take $\log x$ as a constant and use the derivative of the xth power of a constant as $\dfrac{d\left( {{a}^{x}} \right)}{dx}={{a}^{x}}\log a$ because in this formula a is a constant whereas in the question $\log x$ is not a constant but is a function of x.

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