Question

How do you find the derivative of $ y = \ln \ln (2{x^4}) $

Answer 1

Hint: In order to determine the differentiation of the above function with respect to x, we will be using chaining rule by considering $ \ln (2{x^4}) $ as $ f(x) $ and using derivative rule $ \dfrac{d}{{dx}}\ln \left( {f(x)} \right) = \dfrac{1}{{f(x)}}.\dfrac{d}{{dx}}(f(x)) $ .Now applying this chain rule one more time by considering $ 2{x^4} $ as $ g(x) $ . The derivative of variable $ x $ raised to some power n is equal to $ n{x^{n - 1}} $ .Using these properties of the derivative you will get your required answer.

Complete step by step solution:
We are Given a expression $ y = \ln \ln (2{x^4}) $ and we have to find the derivative of this expression with respect to x.
 $ y = \ln \ln (2{x^4}) $
We have to find the first derivative of the above equation
 \[
  \dfrac{d}{{dx}}\left[ y \right] = y' \\
  y' = \dfrac{d}{{dx}}\left( {\ln \ln (2{x^4})} \right) \;
 \]
Let’s assume $ \ln (2{x^4}) $ be a function of x i.e. $ f(x) $ .
 \[y' = \dfrac{d}{{dx}}\left( {\ln (f(x))} \right)\] -----------(1)
Now Applying Chain rule to the above derivative which says that if we are not given a single variable $ x $ and instead of it a function is given( $ f(x) $ )then the derivative will become
 $ \dfrac{d}{{dx}}\ln \left( {f(x)} \right) = \dfrac{1}{{f(x)}}.\dfrac{d}{{dx}}(f(x)) $
We know that Derivative of $ \ln x $ is $ \dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x} $
Putting $ \dfrac{d}{{dx}}\ln \left( {f(x)} \right) = \dfrac{1}{{f(x)}}.\dfrac{d}{{dx}}(f(x)) $ in the equation (1)
 \[y' = \dfrac{1}{{f(x)}}.\dfrac{d}{{dx}}(f(x))\]
Putting back $ f(x) $
 \[y' = \dfrac{1}{{\ln (2{x^4})}}.\dfrac{d}{{dx}}(\ln (2{x^4}))\]
Now applying the chain rule by assuming $ 2{x^4} $ as $ g(x) $ ,
 \[y' = \dfrac{1}{{\ln (2{x^4})}}.\dfrac{d}{{dx}}\ln (g(x))\] --------(2)
so the derivative of $ \ln (g(x)) $ will be defined as
 $ \dfrac{d}{{dx}}\ln \left( {g(x)} \right) = \dfrac{1}{{g(x)}}.\dfrac{d}{{dx}}(g(x)) $
Putting this value in equation (2) we get ,
 \[y' = \dfrac{1}{{\ln (2{x^4})}}.\dfrac{1}{{g(x)}}.\dfrac{d}{{dx}}(g(x))\]
Putting the value of $ g(x) $ ,our equation becomes
 \[y' = \dfrac{1}{{\ln (2{x^4})}}.\dfrac{1}{{2{x^4}}}.\dfrac{d}{{dx}}(2{x^4})\]
As we know the derivative of x raised to power n is $ \dfrac{d}{{dx}}({x^x}) = n{x^{n - 1}} $ ,so
 \[
  y' = \dfrac{1}{{\ln (2{x^4})}}.\dfrac{1}{{2{x^4}}}.\dfrac{d}{{dx}}(2{x^4}) \\
  y' = \dfrac{1}{{\ln (2{x^4})}}.\dfrac{1}{{2{x^4}}}.(4)(2){x^3} \\
  y' = \dfrac{{(4)(2){x^3}}}{{\left( {2{x^4}} \right)\ln (2{x^4})}} \;
 \]
Simplifying more further, we get
 \[y' = \dfrac{4}{{\left( x \right)\ln (2{x^4})}}\]
Therefore , the derivative of $ y = \ln \ln (2{x^4}) $ with respect to x is equal to \[\dfrac{4}{{\left( x \right)\ln (2{x^4})}}\] .
So, the correct answer is “ \[\dfrac{4}{{\left( x \right)\ln (2{x^4})}}\] ”.

Note: 1. Calculus consists of two important concepts one is differentiation and other is integration.
2.What is Differentiation?
It is a method by which we can find the derivative of the function .It is a process through which we can find the instantaneous rate of change in a function based on one of its variables.
Let y = f(x) be a function of x. So the rate of change of $ y $ per unit change in $ x $ is given by:
 $ \dfrac{{dy}}{{dx}} $ .
3. . Indefinite integral=Let $ f(x) $ be a function .Then the family of all its primitives (or antiderivatives) is called the indefinite integral of $ f(x) $ and is denoted by $ \int {f(x)} dx $
The symbol $ \int {f(x)dx} $ is read as the indefinite integral of $ f(x) $ with respect to x.