Question

Find the derivative of $ \dfrac{d}{{dx}}\left( {\ln \left( {\ln {x^2}} \right)} \right) $ .

Answer 1

Hint: We know Chain Rule: $ \dfrac{{df}}{{dx}} = \dfrac{{df}}{{dg}} \times \dfrac{{dg}}{{dh}} \times \dfrac{{dh}}{{dx}} $
By using the Chain rule we can solve this problem.
Since we cannot find the direct derivative of the given question we have to use the chain rule. So we must convert our question in the form of the equation above such that we have to find the values of every term in the above equation and substitute it back. In that way, we would be able to find the solution for the given question.

Complete step by step answer:
Given
 $ \dfrac{d}{{dx}}\left( {\ln \left( {\ln {x^2}} \right)} \right) $
Now to convert the question to perform chain rule the following has to be done:
We know $ f\left( x \right) = \ln \left( {\ln {x^2}} \right)......................\left( i \right) $
So let $ h\left( x \right) = {x^2} $ and $ g\left( x \right) = \ln \left( {h\left( x \right)} \right)..........................\left( {ii} \right) $
Such that we can write $ f\left( x \right) = \ln \left( {g\left( x \right)} \right).......................\left( {iii} \right) $
Now for applying the chain rule we need to find $ \dfrac{{df}}{{dg}} $ , $ \dfrac{{dg}}{{dh}} $ and $ \dfrac{{dh}}{{dx}} $ .
From (i), (ii) and (iii)
 $
   \Rightarrow \dfrac{{df}}{{dg}} = \dfrac{1}{{g\left( x \right)}} \\
   \Rightarrow \dfrac{{dg}}{{dh}} = \dfrac{1}{{h\left( x \right)}} \\
   \Rightarrow \dfrac{{dh}}{{dx}} = 2x \\
  $
Now substituting all the values in the equation of chain rule:
 $
  \dfrac{{df}}{{dx}} = \dfrac{{df}}{{dg}} \times \dfrac{{dg}}{{dh}} \times \dfrac{{dh}}{{dx}} \\
  \;\;\;\;\; = \dfrac{1}{{g\left( x \right)}} \times \dfrac{1}{{h\left( x \right)}} \times 2x \\
  \;\;\;\;\; = \dfrac{1}{{\ln \left( {{x^2}} \right)}} \times \dfrac{1}{{{x^2}}} \times 2x \\
  \dfrac{{df}}{{dx}} = \dfrac{2}{{x\ln {x^2}}}.............................\left( {iv} \right) \\
  $
Also from (i) $ f\left( x \right) = \ln \left( {\ln {x^2}} \right) $
So $ \dfrac{{df}}{{dx}} = \dfrac{d}{{dx}}\ln \left( {\ln \left( {{x^2}} \right)} \right) = \dfrac{2}{{x\ln {x^2}}} $
Now simplifying: \[\dfrac{d}{{dx}}\ln \left( {\ln \left( {{x^2}} \right)} \right) = \dfrac{2}{{x\ln {x^2}}} = \dfrac{2}{{x2\ln x}} = \dfrac{2}{{x\ln x}}..............\left( v \right)\]
 $ \Rightarrow \dfrac{d}{{dx}}\ln \left( {\ln \left( {{x^2}} \right)} \right) = \dfrac{2}{{x\ln x}} $ which is our final answer.

Additional Information:
The Chain Rule can also be written as:
 $ \left( {f(g(x))} \right) = f'(g(x))g'(x) $
It mainly tells us how to differentiate composite functions.

Note:
The chain rule is mainly used for finding the derivative of a composite function.
Also, care must be taken while using the chain rule since it should be applied only on composite functions and applying a chain rule that isn’t composite may result in a wrong derivative.
Some of the logarithmic properties useful for solving questions of derivatives are listed below:
 $
  1.\;{\log _b}\left( {PQ} \right) = {\log _b}\left( P \right) + {\log _b}\left( Q \right) \\
  2.\;{\log _b}\left( {\dfrac{P}{Q}} \right) = {\log _b}\left( P \right) - {\log _b}\left( Q \right) \\
  3.\;{\log _b}\left( {{P^q}} \right) = q \times {\log _b}\left( P \right) \\
  $
Equations ‘1’ ‘2’ and ‘3’ are called Product Rule, Quotient Rule, and Power Rule respectively.