Question

Find the derivative of the following function.
\[y=\sqrt[3]{\dfrac{x\left( {{x}^{2}}+1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}}}\]

Answer 1

Hint: We use the product rule to define the differentiation of products of two or more than two functions and Quotient rule to define the differentiation of two functions as \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v{{u}^{'}}-u{{v}^{'}}}{{{v}^{2}}}\].

Complete step-by-step solution -
We are given the function \[y=\sqrt[3]{\dfrac{x\left( {{x}^{2}}+1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}}}\].
To make the problem a little easier to solve , we will cube both sides of the function .
On cubing both sides of the function , we get
\[{{y}^{3}}=\dfrac{x\left( {{x}^{2}}+1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}}\]
Now, we can see that the RHS of the function is of \[\dfrac{f(x).g(x)}{h(x)}\] type where \[f(x)=x,g(x)={{x}^{2}}+1\] and \[h(x)={{(x-1)}^{2}}\] .
Now , the first thing we will do is to find the derivative of the numerator. It will be helpful in applying quotient rules of differentiation while differentiating the function . To find the derivative of the numerator, we need to apply the product rule of differentiation .
The product rule of differentiation is given by \[\dfrac{d}{dx}\left( f(x).g(x) \right)={{f}^{'}}(x).g(x)+{{g}^{'}}(x).f(x)\]
So, on differentiating the numerator with respect to \[x\] by applying product rule of differentiation , we get
 \[\dfrac{d}{dx}\left( x\left( {{x}^{2}}+1 \right) \right)=1.\left( {{x}^{2}}+1 \right)+x.2x\]
\[={{x}^{2}}+1+2{{x}^{2}}\]
\[=3{{x}^{2}}+1\]
Now, we will differentiate the entire function \[{{y}^{3}}=\dfrac{x\left( {{x}^{2}}+1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}}\] .
To differentiate the function , we need to apply the quotient rule of differentiation, the quotient rule of differentiation is given by \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v{{u}^{'}}-u{{v}^{'}}}{{{v}^{2}}}\].
So ,on differentiating the function by applying quotient rule of differentiation , we get
\[\dfrac{d}{dx}\left( {{y}^{3}} \right)=\dfrac{d}{dx}\left( \dfrac{x\left( {{x}^{2}}+1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}} \right)\]
\[\Rightarrow 3{{y}^{2}}\dfrac{dy}{dx}=\dfrac{{{\left( {{x}^{2}}-1 \right)}^{2}}.\dfrac{d}{dx}\left( x\left( {{x}^{2}}+1 \right) \right)-x\left( {{x}^{2}}+1 \right).\dfrac{d}{dx}\left( {{\left( {{x}^{2}}-1 \right)}^{2}} \right)}{{{\left( {{\left( {{x}^{2}}-1 \right)}^{2}} \right)}^{2}}}\]
\[\Rightarrow 3{{y}^{2}}\dfrac{dy}{dx}=\dfrac{{{\left( {{x}^{2}}-1 \right)}^{2}}.\left( 3{{x}^{2}}+1 \right)-\left( {{x}^{3}}+x \right)\left( 2\left( {{x}^{2}}-1 \right).2x \right)}{{{\left( {{x}^{2}}-1 \right)}^{4}}}\]
Now , we will take \[{{x}^{2}}-1\] common from the numerator.
Taking \[{{x}^{2}}-1\] common from numerator , we get
\[3{{y}^{2}}\dfrac{dy}{dx}=\dfrac{\left( {{x}^{2}}-1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{4}}}\left[ \left( {{x}^{2}}-1 \right).\left( 3{{x}^{2}}+1 \right)-\left( {{x}^{3}}+x \right)\left( 4x \right) \right]\]
\[\Rightarrow 3{{y}^{2}}\dfrac{dy}{dx}=\dfrac{1}{{{\left( {{x}^{2}}-1 \right)}^{3}}}\left[ 3{{x}^{4}}+{{x}^{2}}-3{{x}^{2}}-1-4{{x}^{4}}-4{{x}^{2}} \right]\]
\[\Rightarrow 3{{y}^{2}}\dfrac{dy}{dx}=\dfrac{1}{{{\left( {{x}^{2}}-1 \right)}^{3}}}\left[ -{{x}^{4}}-6{{x}^{2}}-1 \right]\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{3{{y}^{2}}{{\left( {{x}^{2}}-1 \right)}^{3}}}\left[ -{{x}^{4}}-6{{x}^{2}}-1 \right]\] -----\[(i)\]
In the question, it is given
\[y=\sqrt[3]{\dfrac{x\left( {{x}^{2}}+1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}}}\]
So , \[{{y}^{2}}={{\left( \dfrac{x\left( {{x}^{2}}+1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}} \right)}^{\dfrac{2}{3}}}\]
Or , \[\dfrac{1}{{{y}^{2}}}={{\left( \dfrac{{{\left( {{x}^{2}}-1 \right)}^{2}}}{x\left( {{x}^{2}}+1 \right)} \right)}^{\dfrac{2}{3}}}\]
Substituting this value in equation\[(i)\] we get ,
\[\dfrac{dy}{dx}={{\left( \dfrac{{{\left( {{x}^{2}}-1 \right)}^{2}}}{x\left( {{x}^{2}}+1 \right)} \right)}^{\dfrac{2}{3}}}.\dfrac{1}{3{{\left( {{x}^{2}}-1 \right)}^{3}}}\left[ -{{x}^{4}}-6{{x}^{2}}-1 \right]\]
Hence , the derivative of the given function \[y=\sqrt[3]{\dfrac{x\left( {{x}^{2}}+1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}}}\]is given as \[\dfrac{dy}{dx}={{\left( \dfrac{{{\left( {{x}^{2}}-1 \right)}^{2}}}{x\left( {{x}^{2}}+1 \right)} \right)}^{\dfrac{2}{3}}}.\dfrac{1}{3{{\left( {{x}^{2}}-1 \right)}^{3}}}\left[ -{{x}^{4}}-6{{x}^{2}}-1 \right]\]

Note: \[{{\left( {{x}^{2}}-1 \right)}^{2}}\]is a composite function. So , it is of the form \[h(x)=p(q(x))\]. To find its derivative, \[h'(x)=p'(q(x)\times q'(x)\] . So , the derivative of \[{{\left( {{x}^{2}}-1 \right)}^{2}}\] will be \[2\left( {{x}^{2}}-1 \right).2x\] and not \[2\left( {{x}^{2}}-1 \right)\].
Students make such mistakes and end up getting a wrong answer. Such mistakes should be avoided .