Question

What is the derivative of $\dfrac{x}{1+{{x}^{2}}}$?

Answer 1

In the given question, we are asked to find the derivative of the given expression. We can see that the given expression is in the form of a fraction, so in order to find the derivative of the given expression, we will use the quotient rule, which is, $\dfrac{dy}{dx}=\dfrac{u'v-v'u}{{{v}^{2}}}$. Using this above rule, we will differentiate the given expression and find the corresponding value of the given expression. Hence, we will have the derivative of $\dfrac{x}{1+{{x}^{2}}}$.

Complete step by step answer:
According to the given question, we are given an expression which we have to differentiate with respect to ‘x’.
The expression we have is,
$\dfrac{x}{1+{{x}^{2}}}$
First of all, we will let,
$y=\dfrac{x}{1+{{x}^{2}}}$----(1)
We will use the quotient rule to differentiate the given question. The quotient rule is given as follows:
$\dfrac{dy}{dx}=\dfrac{u'v-v'u}{{{v}^{2}}}$
Using the above rule, we will differentiate the given expression and we will get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 1+{{x}^{2}} \right)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
We will now differentiate the terms within the respective brackets having $\dfrac{d}{dx}$ and we will get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 1+{{x}^{2}} \right)(1)-x\left( \dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right) \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
We know that the differentiation of ‘x’ is 1 and the differentiation of $1+{{x}^{2}}$ is carried out, we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 1+{{x}^{2}} \right)(1)-x\left( \left( 0 \right)+\left( 2x \right) \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
Now, we will have the expression simplified further and we get the new expression as,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 1+{{x}^{2}} \right)-x\left( 2x \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
Multiplying the respective terms, we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1+{{x}^{2}}-2{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
In the above expression, the terms in the numerator will get subtracted. And we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1-{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
Therefore, the derivative of the given expression is $\dfrac{dy}{dx}=\dfrac{1-{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$.

Note: The use of the correct formula is very important else the obtained derivative will not be the correct one. Also, while applying the quotient rule, the values should be substituted step wise and the derivative of the functions should be correctly done.
$\dfrac{d}{dx}\left( const. \right)=0$
$\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$ and $\dfrac{d}{dx}\left( x \right)=1$.