How do you differentiate \[f\left( x \right)=\tan x\] twice using the quotient rule?

Hint: This question is from the topic of calculus. In this question, we will use the quotient rule of differentiation. In solving this question, we will first differentiate \[\tan x\] using the quotient rule of differentiation. We will get the new term. After that, we will differentiate that new term using the quotient rule of differentiation. After solving the further question, we will get our answer.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to differentiate the given term twice using the quotient rule.
The given term is
\[f\left( x \right)=\tan x\]
As we know that \[\tan x=\dfrac{\sin x}{\cos x}\], so we can write
\[f\left( x \right)=\dfrac{\sin x}{\cos x}\]
For differentiating the above equation using quotient rule of differentiation, let us first know about the quotient rule.
The formula for quotient rule of differentiation is:
\[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\times \dfrac{du}{dx}-u\times \dfrac{dv}{dx}}{{{v}^{2}}}\]
Where, u and v are functions of any x.
So, for differentiating the equation \[f\left( x \right)=\dfrac{\sin x}{\cos x}\], we can write
\[f'\left( x \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x} \right)\]
\[\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x} \right)\]
Now, using the quotient rule, we can write
\[\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x} \right)=\dfrac{\cos x\times \dfrac{d}{dx}\left( \sin x \right)-\sin x\times \dfrac{d}{dx}\left( \cos x \right)}{{{\left( \cos x \right)}^{2}}}\]
Using the formulas of differentiation that is: \[\dfrac{d}{dx}\left( \sin x \right)=\cos x\] and \[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\], we can write the above equation as
\[\Rightarrow f'\left( x \right)=\dfrac{\cos x\times \cos x-\sin x\times \left( -\sin x \right)}{{{\left( \cos x \right)}^{2}}}\]
The above equation can also be written as
\[\Rightarrow f'\left( x \right)=\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\left( \cos x \right)}^{2}}}\]
Using the formula \[{{\cos }^{2}}x+{{\sin }^{2}}x=1\], we can write the above equation as
\[\Rightarrow f'\left( x \right)=\dfrac{1}{{{\left( \cos x \right)}^{2}}}=\dfrac{1}{{{\cos }^{2}}x}\]
Now, we will again differentiate the above as it is said in the question that we have to differentiate twice, so we can write
\[\Rightarrow \dfrac{d}{dx}\left[ f'\left( x \right) \right]=\dfrac{d}{dx}\left( \dfrac{1}{{{\cos }^{2}}x} \right)\]
We can write the above equation as
\[\Rightarrow f''\left( x \right)=\dfrac{d}{dx}\left( \dfrac{1}{{{\cos }^{2}}x} \right)\]
Now, using the quotient rule of differentiation, we can write
\[\Rightarrow f''\left( x \right)=\dfrac{d}{dx}\left( \dfrac{1}{{{\cos }^{2}}x} \right)=\dfrac{{{\cos }^{2}}x\times \dfrac{d\left( 1 \right)}{dx}-1\times \dfrac{d\left( {{\cos }^{2}}x \right)}{dx}}{{{\left( {{\cos }^{2}}x \right)}^{2}}}\]
Using the formula of differentiation \[\dfrac{d}{dx}\left( 1 \right)=0\], we can write
\[\Rightarrow f''\left( x \right)=\dfrac{{{\cos }^{2}}x\times \left( 0 \right)-\dfrac{d\left( {{\cos }^{2}}x \right)}{dx}}{{{\left( {{\cos }^{2}}x \right)}^{2}}}=\dfrac{0-\dfrac{d\left( {{\cos }^{2}}x \right)}{dx}}{{{\cos }^{4}}x}=-\dfrac{\dfrac{d\left( {{\cos }^{2}}x \right)}{dx}}{{{\cos }^{4}}x}\]
Now, we will use chain rule here. The formula for chain rule is:
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=\dfrac{d}{dx}\left( f'\left( g\left( x \right) \right) \right)g'\left( x \right)\]
Where, ‘f’ and ‘g’ are functions of x.
The above equation can also be written as
\[\Rightarrow f''\left( x \right)=-\dfrac{2\cos x\times \dfrac{d\left( \cos x \right)}{dx}}{{{\cos }^{4}}x}\]
Using the formula \[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\], we can write
\[\Rightarrow f''\left( x \right)=-\dfrac{2\cos x\times \left( -\sin x \right)}{{{\cos }^{4}}x}=\dfrac{2\cos x\times \sin x}{{{\cos }^{4}}x}\]
The above equation can also be written as
\[\Rightarrow f''\left( x \right)=\dfrac{2\cos x\times \sin x}{{{\cos }^{2}}x\times {{\cos }^{2}}x}=\dfrac{2\times \sin x}{{{\cos }^{2}}x\times \cos x}\]
As we know that \[\tan x=\dfrac{\sin x}{\cos x}\], we can write
\[\Rightarrow f''\left( x \right)=\dfrac{2}{{{\cos }^{2}}x}\times \dfrac{\sin x}{\cos x}=\dfrac{2}{{{\cos }^{2}}x}\times \tan x\]
As we know that inverse of ‘cos’ is written as ‘sec’, so we can write
\[\Rightarrow f''\left( x \right)=2{{\sec }^{2}}x\times \tan x\]

Hence, we got that the twice differentiation of \[f\left( x \right)=\tan x\] using quotient rule is: \[f''\left( x \right)=2{{\sec }^{2}}x\tan x\]

Note: We should have a better knowledge in the topic of calculus to solve this type of question easily. We should remember the following formulas to solve this type of question easily:
\[\tan x=\dfrac{\sin x}{\cos x}\]
Quotient rule of differentiation: \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\times \dfrac{du}{dx}-u\times \dfrac{dv}{dx}}{{{v}^{2}}}\]
Differentiation of any constant is always zero. For example: \[\dfrac{d}{dx}\left( 1 \right)=0\]
\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]
\[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\]
\[{{\cos }^{2}}x+{{\sin }^{2}}x=1\]
\[\sec x=\dfrac{1}{\cos x}\]