How do you differentiate \[y={{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{3}}\]?

Hint: We should know the derivatives of some of the functions that includes the functions \[{{x}^{3}}\And {{x}^{2}}\]. The derivatives of these functions are \[3{{x}^{2}}\And 2x\] respectively. To solve the given question, we should know how to differentiate composite functions. The composite functions are functions of the form \[f\left( g(x) \right)\], their derivative is found as, \[\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=\dfrac{d\left( f\left( g(x) \right) \right)}{d\left( g(x) \right)}\dfrac{d\left( g(x) \right)}{dx}\]. We should also know the quotient rule of differentiation which states that \[\dfrac{d\left( \dfrac{f(x)}{g(x)} \right)}{dx}=\dfrac{\dfrac{d\left( f(x) \right)}{dx}g(x)-f(x)\dfrac{d\left( g(x) \right)}{dx}}{{{\left( g(x) \right)}^{2}}}\].

Complete step by step answer:
We know that the derivative of the composite function is evaluated as \[\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=\dfrac{d\left( f\left( g(x) \right) \right)}{d\left( g(x) \right)}\dfrac{d\left( g(x) \right)}{dx}\].
We are given the function \[y={{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{3}}\], we are asked to find its derivative. This is a composite function of the form \[f\left( g(x) \right)\], here we have \[f(x)={{x}^{3}}\And g(x)=\dfrac{{{x}^{2}}+1}{{{x}^{2}}-1}\].
To find the derivative of the given function, we need to find \[\dfrac{d\left( {{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{3}} \right)}{d\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}\], and \[\dfrac{d\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}{dx}\].
We know that the derivative of \[{{x}^{3}}\] with respect to x is \[3{{x}^{2}}\]. Thus, the derivative of \[{{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{3}}\] with respect to \[\dfrac{{{x}^{2}}+1}{{{x}^{2}}-1}\] must be equal to \[3{{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{2}}\]. Hence, we get \[\dfrac{d\left( {{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{3}} \right)}{d\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}=3{{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{2}}\]. Now, we need to find the derivative of \[\dfrac{{{x}^{2}}+1}{{{x}^{2}}-1}\] with respect to x. to do this, we will use the product rule as follows,
\[\dfrac{d\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}{dx}=\dfrac{\dfrac{d\left( {{x}^{2}}+1 \right)}{dx}\left( {{x}^{2}}-1 \right)-\left( {{x}^{2}}+1 \right)\dfrac{d\left( {{x}^{2}}-1 \right)}{dx}}{{{\left( {{x}^{2}}-1 \right)}^{2}}}\]
We know that the derivative of \[{{x}^{2}}\] is \[2x\]. Hence, using it we get
\[\dfrac{d\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}{dx}=\dfrac{2x\left( {{x}^{2}}-1 \right)-\left( {{x}^{2}}+1 \right)2x}{{{\left( {{x}^{2}}-1 \right)}^{2}}}\]
Simplifying the above expression, we get
\[\dfrac{d\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}{dx}=\dfrac{-4x}{{{\left( {{x}^{2}}-1 \right)}^{2}}}\]
Using this, we can evaluate the derivative of the given function as
\[\dfrac{d\left( {{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{3}} \right)}{dx}=\dfrac{d\left( {{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{3}} \right)}{d\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}\times \dfrac{d\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}{dx}\]
\[\dfrac{d\left( {{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{3}} \right)}{dx}=3{{\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \right)}^{2}}\times \dfrac{-4x}{{{\left( {{x}^{2}}-1 \right)}^{2}}}=\dfrac{-12x{{\left( {{x}^{2}}+1 \right)}^{2}}}{{{\left( {{x}^{2}}-1 \right)}^{3}}}\]

Note:
One must know the derivatives of different functions to solve these types of problems. Along with them, we should also know the different properties like product rule and quotient rule for differentiating complex functions.