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.