Question

How do you differentiate $y=\dfrac{{{x}^{3}}}{1-{{x}^{2}}}?$

Answer 1

Hint: We are going to make use of the quotient rule of differentiation. We will find the derivative of the dependent variable $y$ with respect to the independent variable $x.$ The quotient rule is as follows $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ where $u$ and $v$ are two functions of $x.$

Complete step by step solution:
Consider the given equation where $y$ is the dependent variable and $x$ is the independent variable, $y=\dfrac{{{x}^{3}}}{1-{{x}^{2}}}.$
Here, we are asked to find the first derivative of the dependent variable by differentiating the quotient of two functions of $x,$ \[{{x}^{3}}\] and $1-{{x}^{2}}.$
We are going to differentiate this quotient, to find the first derivative of $y,$ using the $\dfrac{u}{v}-$method or the quotient rule, $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ in which the dependent variables $u$ and $v$ are two functions of the independent variable $x.$
Let us compare our problem with the quotient rule so that we can follow the steps of the rule.
Now, we can see that, in the given equation $u={{x}^{3}}$ and $v=1-{{x}^{2}}.$
So, we will get $y=\dfrac{u}{v}.$
Now, the first derivative of $y$ can be obtained by differentiating the given equation once with respect to $x.$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{{{x}^{3}}}{1-{{x}^{2}}} \right).$
Now, we are following the exactly same steps as in the quotient rule,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 1-{{x}^{2}} \right)\dfrac{d{{x}^{3}}}{dx}-{{x}^{3}}\dfrac{d\left( 1-{{x}^{2}} \right)}{dx}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}.$
Now let us find the derivative $\dfrac{d{{x}^{3}}}{dx},$
$\Rightarrow \dfrac{d{{x}^{3}}}{dx}=3{{x}^{2}},$ Since we have $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}.$
Now, consider the derivative $\dfrac{d\left( 1-{{x}^{2}} \right)}{dx},$
$\Rightarrow \dfrac{d\left( 1-{{x}^{2}} \right)}{dx}=\dfrac{d}{dx}1-\dfrac{d}{dx}{{x}^{2}},$ Since $\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}.$
We know that the derivative of a constant is always zero and since $1$ is a constant, $\dfrac{d}{dx}1=0$
$\Rightarrow \dfrac{d\left( 1-{{x}^{2}} \right)}{dx}=0-\dfrac{d{{x}^{2}}}{dx}=-\dfrac{d{{x}^{2}}}{dx}.$
With the help of the identity $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}},$ we will find $\dfrac{d{{x}^{2}}}{dx}=2x$
$\Rightarrow \dfrac{d\left( 1-{{x}^{2}} \right)}{dx}=-2x.$
Therefore, the required derivative will become,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 1-{{x}^{2}} \right)3{{x}^{2}}-{{x}^{3}}\left( -2x \right)}{{{\left( 1-{{x}^{2}} \right)}^{2}}}.$
From this we will get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 3{{x}^{2}}-3{{x}^{4}} \right)+2{{x}^{4}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}.$
Thus, we will get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{3{{x}^{2}}-3{{x}^{4}}+2{{x}^{4}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}.$
So, we will get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{3{{x}^{2}}-{{x}^{4}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}=\dfrac{\left( 3-{{x}^{2}} \right){{x}^{2}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}.$

Hence the required first derivative is $\dfrac{dy}{dx}=\dfrac{\left( 3-{{x}^{2}} \right){{x}^{2}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}.$

Note: The identity $\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}$ is called the linearity property of differentiation. If $u$ and $v$ are two functions of $x,$ then $\dfrac{d}{dx}uv=u\dfrac{dv}{dx}+v\dfrac{du}{dx}.$ This identity is called product rule or the $uv$ method.