How do you differentiate \[\ln \left( \dfrac{x+1}{x-1} \right)\] ?

Hint: This question is from the topic differentiation of calculus chapter. In this question, we are going to differentiate the term \[\ln \left( \dfrac{x+1}{x-1} \right)\]. For solving this question, we are going to use formulas like \[\dfrac{d}{dx}\ln x=\dfrac{1}{x}\] and \[\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{vdu-udv}{{{v}^{2}}}\]. We are going to use chain rule in this question. The chain rule is used to differentiate the composite functions. We will also know about the composite functions in this question.

Complete step by step solution:
Let us solve this question.
The question has asked us to differentiate the term\[\ln \left( \dfrac{x+1}{x-1} \right)\].
The differentiation of \[\ln \left( \dfrac{x+1}{x-1} \right)\] will be
\[\dfrac{d}{dx}\ln \left( \dfrac{x+1}{x-1} \right)\]
For differentiating the above term, we are going to use chain rule here.
The chain rule says that the differentiation of \[f\left( g\left( x \right) \right)\] is \[f'\left( g\left( x \right) \right)\times g'\left( x \right)\]. The chain rule helps us to differentiate composite functions. The composite function should be in the form of f(g(x)), where f(x) and g(x) are two different functions.
We can see in the term \[\ln \left( \dfrac{x+1}{x-1} \right)\]that we have to differentiate is a composite function. Here, g(x) is the function of x that is \[\dfrac{x+1}{x-1}\] and f is the function of \[\ln \] (that is log base e).
So, according to the chain rule, the differentiation of \[\ln \left( \dfrac{x+1}{x-1} \right)\] will be
\[\dfrac{d}{dx}\ln \left( \dfrac{x+1}{x-1} \right)=\dfrac{1}{\dfrac{x+1}{x-1}}\times \dfrac{d}{dx}\left( \dfrac{x+1}{x-1} \right)\]
We have used a formula in the above that is \[\dfrac{d}{dx}\ln x=\dfrac{1}{x}\].
Now, we can write the above equation as
\[\dfrac{d}{dx}\ln \left( \dfrac{x+1}{x-1} \right)=\dfrac{x-1}{x+1}\times \dfrac{d}{dx}\left( \dfrac{x+1}{x-1} \right)\]
Now, we will use the formula\[\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{vdu-udv}{{{v}^{2}}}\], where u and v are two functions of x.
So, we can write the above equation as
\[\Rightarrow \dfrac{d}{dx}\ln \left( \dfrac{x+1}{x-1} \right)=\dfrac{x-1}{x+1}\times \left[ \dfrac{\left( x-1 \right)\dfrac{d}{dx}\left( x+1 \right)-\left( x+1 \right)\dfrac{d}{dx}\left( x-1 \right)}{{{\left( x-1 \right)}^{2}}} \right]\]
The above differentiation can also be written as
\[\Rightarrow \dfrac{d}{dx}\ln \left( \dfrac{x+1}{x-1} \right)=\dfrac{x-1}{x+1}\times \left[ \dfrac{\left( x-1 \right)\left\{ \dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( 1 \right) \right\}-\left( x+1 \right)\left\{ \dfrac{d}{dx}\left( x \right)-\dfrac{d}{dx}\left( 1 \right) \right\}}{{{\left( x-1 \right)}^{2}}} \right]\]
As we know that, the differentiation of x with respect of x is 1 and differentiation of any constant term is always zero. So, we can write the above differentiation as
\[\Rightarrow \dfrac{d}{dx}\ln \left( \dfrac{x+1}{x-1} \right)=\dfrac{x-1}{x+1}\times \left[ \dfrac{\left( x-1 \right)\left\{ 1+0 \right\}-\left( x+1 \right)\left\{ 1-0 \right\}}{{{\left( x-1 \right)}^{2}}} \right]\]
The above can also be written as
\[\Rightarrow \dfrac{d}{dx}\ln \left( \dfrac{x+1}{x-1} \right)=\dfrac{x-1}{x+1}\times \left[ \dfrac{\left( x-1 \right)-\left( x+1 \right)}{{{\left( x-1 \right)}^{2}}} \right]=\dfrac{x-1}{x+1}\times \left[ \dfrac{x-1-x-1}{{{\left( x-1 \right)}^{2}}} \right]=\dfrac{x-1}{x+1}\times \left[ \dfrac{-2}{{{\left( x-1 \right)}^{2}}} \right]\]
The above can also be written as
\[\Rightarrow \dfrac{d}{dx}\ln \left( \dfrac{x+1}{x-1} \right)=\dfrac{x-1}{x+1}\times \left[ \dfrac{-2}{\left( x-1 \right)\left( x-1 \right)} \right]=\dfrac{1}{x+1}\times \dfrac{\left( -2 \right)}{\left( x-1 \right)}\]
The above can also be written as
\[\Rightarrow \dfrac{d}{dx}\ln \left( \dfrac{x+1}{x-1} \right)=\dfrac{-2}{\left( x+1 \right)\left( x-1 \right)}\]

Hence, we get that differentiation of \[\ln \left( \dfrac{x+1}{x-1} \right)\] is \[\dfrac{-2}{\left( x+1 \right)\left( x-1 \right)}\].

Note: We should have a better knowledge in the topic of differentiation to solve this type of question easily. The formulas we have used here in the solution are:
\[\dfrac{d}{dx}\ln x=\dfrac{1}{x}\];
\[\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{vdu-udv}{{{v}^{2}}}\], where u and v are two functions of x
Don’t forget the above formulas to solve this type of question easily.
And, also remember the chain rule. The chain rule helps us to differentiate the composite functions like\[f\left( g\left( x \right) \right)\]. So, \[\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\times g'\left( x \right)\]. Here, f and g are two different functions, and \[f'\] and \[g'\] are differentiation of f and g respectively.