How do you differentiate the function $f(x)=\dfrac{{{x}^{2}}+2x}{{{x}^{2}}-4}$?

Hint: In this question we have the given function in the form of a fraction therefore, we will use the form of derivative of $\dfrac{u}{v}$ which is $\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$. we will also use the formula of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the denominator of the expression. We will then multiply and simplify the terms to get the required solution.

Complete step-by-step solution:
We have the given expression as:
$\Rightarrow f(x)=\dfrac{{{x}^{2}}+2x}{{{x}^{2}}-4}$
We have to find the derivative of the expression therefore; we can write as:
$\Rightarrow f'(x)=\dfrac{d}{dx}\dfrac{{{x}^{2}}+2x}{{{x}^{2}}-4}$
Now since it is in the form of $\dfrac{u}{v}$, we will use the formula $\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ by considering $u={{x}^{2}}+2x$ and $v={{x}^{2}}-4$.
On using the formula, we get:
\[\Rightarrow f'(x)=\dfrac{\left( {{x}^{2}}-4 \right)\dfrac{d}{dx}\left( {{x}^{2}}+2x \right)-\left( {{x}^{2}}+2x \right)\dfrac{d}{dx}\left( {{x}^{2}}-4 \right)}{{{\left( {{x}^{2}}-4 \right)}^{2}}}\]
Now the term ${{x}^{2}}-4$ can be written as ${{x}^{2}}-{{2}^{2}}$ and since it is in the form of ${{a}^{2}}-{{b}^{2}}$, we will expand it as $\left( a+b \right)\left( a-b \right)$.
\[\Rightarrow f'(x)=\dfrac{\left( {{x}^{2}}-4 \right)\dfrac{d}{dx}\left( {{x}^{2}}+2x \right)-\left( {{x}^{2}}+2x \right)\dfrac{d}{dx}\left( {{x}^{2}}-4 \right)}{{{\left( \left( x-2 \right)\left( x+2 \right) \right)}^{2}}}\]
On splitting the square in the denominator, we get:
\[\Rightarrow f'(x)=\dfrac{\left( {{x}^{2}}-4 \right)\dfrac{d}{dx}\left( {{x}^{2}}+2x \right)-\left( {{x}^{2}}+2x \right)\dfrac{d}{dx}\left( {{x}^{2}}-4 \right)}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}\]
Now we know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ therefore $\dfrac{d}{dx}{{x}^{2}}=2x$ and $\dfrac{dx}{dx}=1$ . we also know that $\dfrac{d}{dx}k=0$ therefore on using these formulas and substituting, we get:
\[\Rightarrow f'(x)=\dfrac{\left( {{x}^{2}}-4 \right)\left( 2x+2 \right)-\left( {{x}^{2}}+2x \right)\left( 2x \right)}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}\]
On multiplying the terms, we get:
\[\Rightarrow f'(x)=\dfrac{\left( 2{{x}^{3}}+2{{x}^{2}}-8x-8 \right)-\left( 2{{x}^{3}}+4{{x}^{2}} \right)}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}\]
On removing the brackets, we get:
\[\Rightarrow f'(x)=\dfrac{2{{x}^{3}}+2{{x}^{2}}-8x-8-2{{x}^{3}}-4{{x}^{2}}}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}\]
On simplifying, we get:
\[\Rightarrow f'(x)=\dfrac{2{{x}^{2}}-8x-8-4{{x}^{2}}}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}\]
On further simplifying, we get:
\[\Rightarrow f'(x)=\dfrac{-2{{x}^{2}}-8x-8}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}\]
Now on taking $-2$ common from the numerator, we get:
\[\Rightarrow f'(x)=\dfrac{-2\left( {{x}^{2}}+4x+4 \right)}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}\]
Now the term ${{x}^{2}}+4x+4$ is in the form of the expansion of ${{\left( x+2 \right)}^{2}}$, on substituting, we get:
\[\Rightarrow f'(x)=\dfrac{-2{{\left( x+2 \right)}^{2}}}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}\]
On cancelling the terms, we get:
\[\Rightarrow f'(x)=\dfrac{-2}{{{\left( x-2 \right)}^{2}}}\], which is the required solution.

Note: In this question we have used the $\dfrac{d}{dx}\dfrac{u}{v}$ formula which is for two terms which are in division. There also exists the formula for two terms in division which can be denoted as $\dfrac{d}{dx}uv$ and the formula is written as $\dfrac{d}{dx}uv=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. It is to be also remembered that differentiation is the reverse of integration.