# Differentiate the given function: g(x) = \frac {(x^2+x+1)(4-x)}{2x-1}

## Question:

Differentiate the given function: {eq}g(x) = \frac {(x^2+x+1)(4-x)}{2x-1} {/eq}

## Quotient rule of Differentiation:

When the function to be differentiated is in the form of a fraction, then its differentiation can not be found out simply by calculating the derivative of numerator and denominator. Instead of, differentiation has to be carried out with the help of the quotient rule of differentiation. If F(x) is a function such as:

{eq}F(x) = \dfrac {f(x)}{g(x)} {/eq}

then,

{eq}F'(x) = \dfrac {g(x) * f'(x) - f(x) * g'(x)}{[g(x)]^2} {/eq}

where f(x) and g(x) are two differentiable functions.

Given:

{eq}\displaystyle { g(x) = \frac {(x^2+x+1)(4-x)}{2x-1} } {/eq}

Simplifying the given function:

{eq}\displaystyle { g(x) = \frac {4x^2+ 4x+ 4 - x^3 - x^2 - x}{2x-1} \\ \Rightarrow g(x) = \frac {- x^3 + 3x^2 + 3x + 4}{2x-1} \\ } {/eq}

Differentiating the above function:

{eq}\displaystyle { g'(x) = \dfrac { (2x-1) (- x^3 + 3x^2 + 3x + 4)' - (- x^3 + 3x^2 + 3x + 4) (2x-1)' }{ (2x-1)^2} \\ \Rightarrow g'(x) = \dfrac { (2x-1) (- 3x^2 + 3 (2x) + 3) - (- x^3 + 3x^2 + 3x + 4) (2) }{ (2x-1)^2} \\ \Rightarrow g'(x) = \dfrac { (2x-1) (- 3x^2 + 6x + 3) - (- 2x^3 + 6x^2 + 6x + 8) }{ (2x-1)^2} \\ \Rightarrow g'(x) = \dfrac { (-6x^3 + 12x^2 + 6x + 3x^2 - 6x - 3) - (- 2x^3 + 6x^2 + 6x + 8) }{ (2x-1)^2} \\ \Rightarrow g'(x) = \dfrac { -6x^3 + 15x^2 - 3 + 2x^3 - 6x^2 - 6x - 8) }{ (2x-1)^2} \\ \Rightarrow g'(x) = \dfrac { -4x^3 + 9x^2 - 6x - 11 }{ (2x-1)^2} \\ } {/eq}