Question

If $y = \log .\log x$ then ${e^y}\dfrac{{dy}}{{dx}}$ is equals to
A. $\dfrac{1}{x}$
B. $\dfrac{1}{{x\log x}}$
C. $\dfrac{1}{{\log x}}$
D. ${e^y}$

Answer 1

Given is the function with logarithmic terms. we have to find its first order derivative or simply we can say derivative. For that we should know the derivative of log x with the help of chain rule here. After that we can take the original function in exponential form and then take the product of it with the derivative which is our answer.

Complete step by step answer:
Given is the function,
$y = \log .\log x$
Here we know that, $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
Then we can write,
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}}\dfrac{1}{x}$
Now for the original function we will take exponential form on both sides,
${e^y} = {e^{\log .\log x}}$
We can say that, ${e^{\log .\log x}} = \log x$
${e^y} = \log x$
Now taking the product of the terms,
${e^y}\dfrac{{dy}}{{dx}} = \log x.\dfrac{1}{{\log x}}\dfrac{1}{x}$
On cancelling the logx term,
$\therefore {e^y}\dfrac{{dy}}{{dx}} = \dfrac{1}{x}$
Thus this is the correct answer.

So option A is correct.

Note: Finding the derivative of a function is a mathematical process. There are different functions and rules that are followed when we find the derivative. Here students get confused on a very initial step of finding the derivative. So what we have done is we have used chain rule in which there are cumulative functions involved in between. This is the only thing to note.