Question

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

Answer 1

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 :
$\begin{aligned} & \Rightarrow \dfrac{d}{dx}\left( 2x \right)=2......eqn\left( 1 \right) \\\ & \Rightarrow \dfrac{d}{dx}\left( 1+{{x}^{2}} \right)=\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right)=0+2x.....eqn\left( 2 \right) \\\ & \\\ & \\\ \end{aligned}$
$\Rightarrow \dfrac{d}{dx}\left( k \right)=0$ where k is any random constant.
Now , let’s differentiate and find out the derivative of the function using the above mentioned rule.
$\begin{aligned} & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{\left( 1+{{x}^{2}} \right)\dfrac{d}{dx}\left( 2x \right)-\left( 2x \right)\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\\ & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{\left( 1+{{x}^{2}} \right)\left( 2 \right)-\left( 2x \right)\left( 2x \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\\ & \\\ \end{aligned}$
From $eqn\left( 1 \right)\And eqn\left( 2 \right)$ :
$\begin{aligned} & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2+2{{x}^{2}}-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\\ & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2-2{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\\ \end{aligned}$
$\dfrac{d}{dx}\left( f\left( x \right) \right)$ 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.