Question

# Differentiate ${{\left( \log x \right)}^{x}}$ with respect to $\log x$ and obtain the answer.

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.

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.