Question

How do differentiate $y=\ln \left( \ln \left( 2{{x}^{4}} \right) \right)$?

Answer 1

For the given question it is given to derivate $y=\ln \left( \ln \left( 2{{x}^{4}} \right) \right)$. For that we have to consider it as equation (1) and then differentiate the problem using chain rule, we have to use chain rule two times to solve this problem and at the end of the problem we have to use product rule for the final answer.

Complete step-by-step solution:
To find the derivative for the given equation$y=\ln \left( \ln \left( 2{{x}^{4}} \right) \right)$. Let us consider the given equation as equation (1).
Considering the above equation as equation (1).
$y=\ln \left( \ln \left( 2{{x}^{4}} \right) \right).........\left( 1 \right)$
As we know the chain rule is$\dfrac{d}{dx}\ln \left( f\left( x \right) \right)=\dfrac{1}{f\left( x \right)}.\dfrac{d}{dx}f\left( x \right)$.
Let us consider the above formula as formula (f1).
$\dfrac{d}{dx}\ln \left( f\left( x \right) \right)=\dfrac{1}{f\left( x \right)}.\dfrac{d}{dx}f\left( x \right)........\left( f1 \right)$.
By applying the formula (f1) in equation (1), we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}.\ln \left( \ln \left( 2{{x}^{4}} \right) \right)$
$\Rightarrow \dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\dfrac{d}{dx}\ln \left( 2{{x}^{4}} \right)$
Let us consider the above equation as equation (2).
As we can see that we can also apply formula (f1) for the equation (2) , again we get the function in the same form.
$\dfrac{dy}{dx}=\dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\dfrac{d}{dx}\ln \left( 2{{x}^{4}} \right)...........\left( 2 \right)$
By applying the formula (f1) with equation (2) we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\left[ \dfrac{1}{2{{x}^{4}}}.\dfrac{d}{dx}.2{{x}^{4}} \right]$
Let us consider the above equation as equation (3).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\left[ \dfrac{1}{2{{x}^{4}}}.\dfrac{d}{dx}.2{{x}^{4}} \right].........\left( 3 \right)$
As we can see the term $\dfrac{d}{dx}.2{{x}^{4}}$ we can apply the power rule to find the derivative of the term.
Therefore the power rule is$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$.
Let us consider the above formula as (f2).
$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}..........(f2)$
So, by applying the formula (f2) to the equation (3) we get
$\Rightarrow \dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\left[ \dfrac{1}{2{{x}^{4}}}.8{{x}^{3}} \right]$
By simplifying a bit, we get
$\Rightarrow \dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\left[ \dfrac{4}{x} \right]$
Let us consider the given equation as equation (4).
$\dfrac{dy}{dx}=\dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\left[ \dfrac{4}{x} \right]...........\left( 4 \right)$
Therefore equation (4) i.e. $\dfrac{dy}{dx}=\dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\left[ \dfrac{4}{x} \right]$is the solution for equation (1) i.e. $y=\ln \left( \ln \left( 2{{x}^{4}} \right) \right)$.

Note: While solving this problems students may struck at $\dfrac{dy}{dx}=\dfrac{d}{dx}.\ln \left( \ln \left( 2{{x}^{4}} \right) \right)$ because from the formula we can see that there is no log in solution. So, students may confused and write the solution as $\dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\dfrac{d}{dx}\left( 2{{x}^{4}} \right)$ . But we have to understand that in our question it is given $\log \left( \log \left( {} \right) \right)$ so the solution of the equation will be $\dfrac{1}{\ln \left( 2{{x}^{4}} \right)}.\dfrac{d}{dx}\ln \left( 2{{x}^{4}} \right)$.