Solveeit Logo
Login

Question

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

How do you differentiate the function f(x)=x2+2xx2−4f(x)=\dfrac{{{x}^{2}}+2x}{{{x}^{2}}-4}?

Explanation

Solution

In this question we have the given function in the form of a fraction therefore, we will use the form of derivative of uv\dfrac{u}{v} which is ddxuv=vdudx−udvdxv2\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}. we will also use the formula of a2−b2=(a+b)(a−b){{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:
⇒f(x)=x2+2xx2−4\Rightarrow f(x)=\dfrac{{{x}^{2}}+2x}{{{x}^{2}}-4}
We have to find the derivative of the expression therefore; we can write as:
⇒f′(x)=ddxx2+2xx2−4\Rightarrow f'(x)=\dfrac{d}{dx}\dfrac{{{x}^{2}}+2x}{{{x}^{2}}-4}
Now since it is in the form of uv\dfrac{u}{v}, we will use the formula ddxuv=vdudx−udvdxv2\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}} by considering u=x2+2xu={{x}^{2}}+2x and v=x2−4v={{x}^{2}}-4.
On using the formula, we get:
⇒f′(x)=(x2−4)ddx(x2+2x)−(x2+2x)ddx(x2−4)(x2−4)2\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 x2−4{{x}^{2}}-4 can be written as x2−22{{x}^{2}}-{{2}^{2}} and since it is in the form of a2−b2{{a}^{2}}-{{b}^{2}}, we will expand it as (a+b)(a−b)\left( a+b \right)\left( a-b \right).
⇒f′(x)=(x2−4)ddx(x2+2x)−(x2+2x)ddx(x2−4)((x−2)(x+2))2\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:
⇒f′(x)=(x2−4)ddx(x2+2x)−(x2+2x)ddx(x2−4)(x−2)2(x+2)2\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 ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} therefore ddxx2=2x\dfrac{d}{dx}{{x}^{2}}=2x and dxdx=1\dfrac{dx}{dx}=1 . we also know that ddxk=0\dfrac{d}{dx}k=0 therefore on using these formulas and substituting, we get:
⇒f′(x)=(x2−4)(2x+2)−(x2+2x)(2x)(x−2)2(x+2)2\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:
⇒f′(x)=(2x3+2x2−8x−8)−(2x3+4x2)(x−2)2(x+2)2\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:
⇒f′(x)=2x3+2x2−8x−8−2x3−4x2(x−2)2(x+2)2\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:
⇒f′(x)=2x2−8x−8−4x2(x−2)2(x+2)2\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:
⇒f′(x)=−2x2−8x−8(x−2)2(x+2)2\Rightarrow f'(x)=\dfrac{-2{{x}^{2}}-8x-8}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}
Now on taking −2-2 common from the numerator, we get:
⇒f′(x)=−2(x2+4x+4)(x−2)2(x+2)2\Rightarrow f'(x)=\dfrac{-2\left( {{x}^{2}}+4x+4 \right)}{{{\left( x-2 \right)}^{2}}{{\left( x+2 \right)}^{2}}}
Now the term x2+4x+4{{x}^{2}}+4x+4 is in the form of the expansion of (x+2)2{{\left( x+2 \right)}^{2}}, on substituting, we get:
⇒f′(x)=−2(x+2)2(x−2)2(x+2)2\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:
⇒f′(x)=−2(x−2)2\Rightarrow f'(x)=\dfrac{-2}{{{\left( x-2 \right)}^{2}}}, which is the required solution.

Note: In this question we have used the ddxuv\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 ddxuv\dfrac{d}{dx}uv and the formula is written as ddxuv=udvdx+vdudx\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.