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**Hint:**Derivative is the rate of change of a function. To find the derivative of a function, we have to differentiate it with respect to $x$. Finding out the derivative of a function implies that we are finding out the slope of the function. And for this particular function we need to make use of the $\dfrac{u}{v}$ rule of differentiation. Whenever we have two functions namely, $h\left( x \right),g\left( x \right)$in this form $\dfrac{h\left( x \right)}{g\left( x \right)}$ , then we differentiate such a function using this rule. It states the following :

$\dfrac{d}{dx}\left( \dfrac{h\left( x \right)}{g\left( x \right)} \right)=\dfrac{g\left( x \right)\dfrac{d}{dx}\left( h\left( x \right) \right)-h\left( x \right)\dfrac{d}{dx}\left( g\left( x \right) \right)}{{{\left( g\left( x \right) \right)}^{2}}}$ .

**Complete step-by-step answer:**

We know that $f\left( x \right)=\dfrac{2x}{1+{{x}^{2}}}=\dfrac{h\left( x \right)}{g\left( x \right)}$.

Upon comparing $h\left( x \right)=2x$ and $g\left( x \right)=1+{{x}^{2}}$.

We know the following :

$\begin{align}

& \Rightarrow \dfrac{d}{dx}\left( 2x \right)=2......eqn\left( 1 \right) \\

& \Rightarrow \dfrac{d}{dx}\left( 1+{{x}^{2}} \right)=\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right)=0+2x.....eqn\left( 2 \right) \\

& \\

& \\

\end{align}$

$\Rightarrow \dfrac{d}{dx}\left( k \right)=0$ where k is any random constant.

Now , let’s differentiate and find out the derivative of the function using the above mentioned rule.

$\begin{align}

& \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{\left( 1+{{x}^{2}} \right)\dfrac{d}{dx}\left( 2x \right)-\left( 2x \right)\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\

& \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{\left( 1+{{x}^{2}} \right)\left( 2 \right)-\left( 2x \right)\left( 2x \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\

& \\

\end{align}$

From $eqn\left( 1 \right)\And eqn\left( 2 \right)$ :

$\begin{align}

& \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2+2{{x}^{2}}-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\

& \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2-2{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\

\end{align}$

$\dfrac{d}{dx}\left( f\left( x \right) \right)$ can also be represented as $f'\left( x \right)$ .

Now , we found the first derivative. Using this first derivative , we can find out where a particular function reaches its maximum value or minimum value and where it just stays constant. This is specifically called the first derivative test. This test helps us a lot in tracing the particular function.

There’s also another way to find out the derivative of this function. We can substitute$x$. We have to plug-in $x=\tan \theta $ .

We know that :

$\sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }$

We can make use of this information and differentiate it and we will end up with the same value.

$\therefore f'\left( x \right)=\dfrac{2-2{{x}^{2}}}{1+{{x}^{2}}}$

**Note:**We have to be careful while applying the formula. We should carefully compare and find out what $h\left( x \right)\And g\left( x \right)$ are . It is advisable to learn the derivatives of all the standard functions in order to solve the question quickly. And if we want to do this question by making use of $x=\tan \theta $ , then we have to make sure that we are converting everything into $x$ at the end as we used $x=\tan \theta $ for our convenience.

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