Question

How do you differentiate the function $\dfrac{x\sin x}{{{x}^{2}}+1}$?

Answer 1

Hint: Now we are given with a function in fractional form. Hence to differentiate the function
Now the given function is in division form. Hence we will use the division rule to solve the differentiation. Now we have differentiation of $\dfrac{f\left( x \right)}{g\left( x \right)}$ is given by the formula $\dfrac{f'\left( x \right)g\left( x \right)-g'\left( x \right)f\left( x \right)}{{{g}^{2}}\left( x \right)}$ . Now we will differentiate $x\sin x$ by using chain rule and ${{x}^{2}}+1$ by using differentiation of $\dfrac{d\left( f+g \right)}{dx}=\dfrac{df}{dx}+\dfrac{dg}{dx}$ . Hence we will substitute the values in the formula and then find the value of the required differentiation.

Complete step-by-step answer:
Now to solve the differentiation we will use the division rule of differentiation.
Now we have differentiation of the function of the form $\dfrac{f\left( x \right)}{g\left( x \right)}$ is given by the formula $\dfrac{f'\left( x \right)g\left( x \right)-g'\left( x \right)f\left( x \right)}{{{g}^{2}}\left( x \right)}$ .
Now here we have $f\left( x \right)=x\sin x$ and $g\left( x \right)={{x}^{2}}+1$
Now first consider $f\left( x \right)=x\sin x$ .
Now to differentiate the equation we will use product rule of differentiation.
Now according to product rule of differentiation we have, differentiation of the function $f\left( x \right).g\left( x \right)$ is given by $f'\left( x \right)g\left( x \right)+g'\left( x \right)f\left( x \right)$. Hence differentiating the function $f\left( x \right)=x\sin x$ we get,
$\Rightarrow f'\left( x \right)=x\cos x+\sin x$
Now consider the function $g\left( x \right)={{x}^{2}}+1$
Now we know that $\dfrac{d\left( f+g \right)}{dx}=\dfrac{df}{dx}+\dfrac{dg}{dx}$ and $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ .
Also we know that differentiation of constant is 0.
Now we know that $g\left( x \right)={{x}^{2}}+1$ , Hence differentiating the function we get,
$\Rightarrow g'\left( x \right)=2x$
Now substituting the value of the terms in the formula we get,
\[\Rightarrow \dfrac{\left( x\cos x+\sin x \right)\left( {{x}^{2}}+1 \right)-\left( 2x \right)\left( x\sin x \right)}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\]
Now simplifying the above expression get
$\Rightarrow \dfrac{\left( {{x}^{2}}+1 \right)\left( x\cos x \right)+\left( {{x}^{2}}+1 \right)\sin x-2{{x}^{2}}\sin x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}$
Now using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$


$\Rightarrow \dfrac{{{x}^{3}}\cos x+x\cos x+{{x}^{2}}\sin x+\sin x-2{{x}^{2}}\sin x}{{{x}^{4}}+2{{x}^{2}}+1}$
Hence the differentiation of the given expression is $\dfrac{{{x}^{3}}\cos x+x\cos x-{{x}^{2}}\sin x+\sin x}{{{x}^{4}}+2{{x}^{2}}+1}$

Note: Now note that while differentiating the function we can also use the product rule by considering $f\left( x \right)\times \dfrac{1}{g\left( x \right)}$ and then differentiating $\dfrac{1}{g\left( x \right)}$ by writing ${{\left( g\left( x \right) \right)}^{-1}}$ and using chain rule of differentiation. Hence we can solve the given problem and differentiate.