Question

How do you differentiate \[f(x) = \log (\log x)\].

Answer 1

Hint: In calculus, differentiation is one of the two important concepts apart from integration. Differentiation is a method of finding the derivative of a function. Differentiation is a process, in Maths, where we find the instantaneous rate of change in function based on one of its variables. The most common example is the rate change of displacement with respect to time, called velocity. The opposite of finding a derivative is anti-differentiation.
If x is a variable and y is another variable, then the rate of change of x with respect to y is given by dy/dx. This is the general expression of the derivative of a function and is represented as f'(x) = dy/dx, where y = f(x) is any function.

Complete step by step answer:
For solving this we need to know the derivative of $\log (x)$ so the derivative of it will be:
$ \Rightarrow \dfrac{d}{{dx}}\log (x) = \dfrac{1}{x}$
As we know the derivative of $\log (x)$ so for outer log the whole inner part will be like x so when we will took the derivative:
$ \Rightarrow f(x) = \log (\log x)$
After differentiating it:
$ \Rightarrow f'(x) = \dfrac{1}{{\log (x)}}\dfrac{d}{{dx}}(\log x)$
Now on further differentiating the inner part of the logarithm we will get:
$ \Rightarrow f'(x) = \dfrac{1}{{\log (x)}}\dfrac{1}{x}$
So this will be our final answer.

Note: Here are some basic derivatives of different functions:
$ \Rightarrow \dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$ \Rightarrow \dfrac{d}{{dx}}(\sin x) = \cos x$
$ \Rightarrow \dfrac{d}{{dx}}(\cos x) = - \sin x$
$ \Rightarrow \dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$
$ \Rightarrow \dfrac{d}{{dx}}(\cot x) = - \cos e{c^2}x$
$ \Rightarrow \dfrac{d}{{dx}}(\sec x) = \sec x\tan x$
$ \Rightarrow \dfrac{d}{{dx}}\cos ecx = - \cos ecx.\cot x$
With the help of this we can solve plenty of simple and difficult questions.