Question

How do you differentiate $\dfrac{1}{\left( 1+{{x}^{2}} \right)}$ ?

Answer 1

Hint: From the given question we have to find the differentiation of $\dfrac{1}{\left( 1+{{x}^{2}} \right)}$. This can be done by quotient rule. The quotient rule is $\dfrac{d}{dx}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{{{f}^{|}}\left( x \right)\times g\left( x \right)-{{g}^{|}}\left( x \right)\times f\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}$ .Here $f\left( x \right)=1$ and $g\left( x \right)=1+{{x}^{2}}$. By this we can differentiate the above $\dfrac{1}{\left( 1+{{x}^{2}} \right)}$ as we know that differentiation of any constant is zero and differentiation of $1+{{x}^{2}}$ is $2x$. Hence, we will get the required answer.

Complete step-by-step solution:
From the question given we have to differentiate the $\dfrac{1}{\left( 1+{{x}^{2}} \right)}$.
Now we have to differentiate this by using the quotient rule in differentiation. The quotient rule is if any differentiable function is in the form of $\dfrac{f\left( x \right)}{g\left( x \right)}$ then we will use the quotient rule. The differentiation of $\dfrac{f\left( x \right)}{g\left( x \right)}$ by quotient rule is,
  $\dfrac{d}{dx}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{{{f}^{|}}\left( x \right)\times g\left( x \right)-{{g}^{|}}\left( x \right)\times f\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}$
From the question given in place of $f\left( x \right)$there is $1$ and in place of $g\left( x \right)$ there is $1+{{x}^{2}}$ i.e., here
$\Rightarrow f\left( x \right)=1$
$\Rightarrow g\left( x \right)=1+{{x}^{2}}$
By substituting in the formula, we will get,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{1+{{x}^{2}}} \right)=\dfrac{\dfrac{d}{dx}\left( 1 \right)\times \left( 1+{{x}^{2}} \right)-\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)\times 1}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
We know that the differentiation of any constant is zero so, the $\dfrac{d}{dx}\left( 1 \right)=0$
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{1+{{x}^{2}}} \right)=\dfrac{0\times \left( 1+{{x}^{2}} \right)-\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)\times 1}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
Now differentiation of $\left( 1+{{x}^{2}} \right)$ is $2x$. Because we know that differentiation of $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ by this we can say that differentiation of $\left( 1+{{x}^{2}} \right)$ is $2x$. Already we know that differentiation of constant is zero so the differentiation of one is zero.
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{1+{{x}^{2}}} \right)=\dfrac{0\times \left( 1+{{x}^{2}} \right)-\left( 0+2x \right)\times 1}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
by simplifying the above expression further, we will get,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{1+{{x}^{2}}} \right)=\dfrac{-\left( 2x \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
Therefore, the differentiation of the $\dfrac{1}{\left( 1+{{x}^{2}} \right)}$ is $\dfrac{-\left( 2x \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$.

Note: Students should know the basic formulas of differentiation. Student should recall all the formulas of differentiation while doing the problem. This problem can be solve by using chain rule also, first rewrite the $\dfrac{1}{\left( 1+{{x}^{2}} \right)}$ as ${{\left( 1+{{x}^{2}} \right)}^{-1}}$ and then we have to differentiate this. It is in the form of $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ so will get $\dfrac{-\left( 2x \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$.