Question

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

Answer 1

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 ${(\log x)}^x$ with respect to $\log x$ i.e. we have to find ${\displaystyle \frac{d{(\log x)}^x}{d(\log x)}}$.

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

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

Therefore, taking $u={(\log x)}^x$ and $v=\log x$ in equation (1.1), we obtain
${\displaystyle \frac{d{(\log x)}^x}{d(\log x)}}={\displaystyle \frac{{\displaystyle \frac{d{(\log x)}^x}{dx}}}{{\displaystyle \frac{d(\log x)}{dx}}}}={\displaystyle \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}}................(1.2)$

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

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

Now, the derivative of log function is given by

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

The chain rules is stated as

${\displaystyle \frac{d(f(g(x)))}{dx}}={\displaystyle \frac{df(g)}{dg}}\times {\displaystyle \frac{dg(x)}{dx}}........(1.5)$


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

${\displaystyle \frac{d(f(x)g(x))}{dx}}=g(x){\displaystyle \frac{df(x)}{dx}}+f(x){\displaystyle \frac{dg(x)}{dx}}...........(1.6)$

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

$$\begin{array}{rl}& \log u=x\log (\log x)\\ & \Rightarrow {\displaystyle \frac{d\log u}{dx}}={\displaystyle \frac{d(x\log (\log x))}{dx}}\\ & \Rightarrow {\displaystyle \frac{d\log u}{du}}{\displaystyle \frac{du}{dx}}=\log (\log x){\displaystyle \frac{d\left(x\right)}{dx}}+x{\displaystyle \frac{d(\log (\log x))}{dx}}\\ & \Rightarrow {\displaystyle \frac{1}{u}}{\displaystyle \frac{du}{dx}}=\log (\log x)\times 1+x\times {\displaystyle \frac{d(\log (\log x))}{d\log x}}{\displaystyle \frac{d\log x}{dx}}\\ & \Rightarrow {\displaystyle \frac{1}{u}}{\displaystyle \frac{du}{dx}}=\log (\log x)+x\times {\displaystyle \frac{1}{\log x}}\times {\displaystyle \frac{1}{x}}\\ & \Rightarrow {\displaystyle \frac{du}{dx}}=u(\log (\log x)+{\displaystyle \frac{1}{\log x}})={(\log x)}^x(\log (\log x)+{\displaystyle \frac{1}{\log x}}).....(1.7)\end{array}$$

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

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

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

$\begin{array}{rl}& {\displaystyle \frac{d{(\log x)}^x}{d(\log x)}}={\displaystyle \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}}={\displaystyle \frac{{(\log x)}^x(\log (\log x)+{\displaystyle \frac{1}{\log x}})}{{\displaystyle \frac{1}{x}}}}\\ & \Rightarrow {\displaystyle \frac{d{(\log x)}^x}{d(\log x)}}=x{(\log x)}^x(\log (\log x)+{\displaystyle \frac{1}{\log x}})\end{array}$

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 ${\displaystyle \frac{d\left(a^x\right)}{dx}}=a^x\log a$ because in this formula a is a constant whereas in the question $\log x$ is not a constant but is a function of x.