# y = \frac {x^{2} + x}{x - 1} [(x - 1) (2x)] - [(x^{2} + x) (1] 2x^{2} - 2x - x^{2} - x where is...

## Question:

{eq}y = \frac {x^{2} + x}{x - 1} {/eq}

{eq}[(x - 1) (2x)] - [(x^{2} + x) (1] {/eq}

{eq}2x^{2} - 2x - x^{2} - x {/eq}

where is the -1 coming from?

{eq}\frac{x^{2} - 3x}{(x - 1)^{2}} {/eq}

## Quotient Rule

To find the derivative of a quotient of functions, the quotient rule must be used. The quotient rule states that {eq}\left(\dfrac{f(x)}{g(x)}\right)' = \dfrac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}. {/eq} This means that the derivative of a quotient can be found by first finding the derivative of the function in the numerator and the derivative of the function in the denominator, and then filling in the appropriate functions and derivatives into the rule.

To find the derivative of {eq}y=\dfrac{x^2+x}{x-1} {/eq} we need to use the quotient rule where {eq}f(x) = x^2 + x {/eq} and {eq}g(x) = x-1. {/eq} Remember that the quotient rule states that {eq}\left(\dfrac{f(x)}{g(x)}\right)' = \dfrac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}. {/eq}

Let's start by finding the derivatives of the functions we defined above. We have

{eq}f(x) = x^2 + x\\ f'(x) = 2x + 1 {/eq} and {eq}g(x) = x-1\\ g'(x) = 1. {/eq}

Substituting this information into the numerator of the complicated expression for the quotient rule, we have

{eq}g(x)f'(x) - f(x)g'(x) = (x-1)(2x+1) - (x^2+ x)(1) {/eq}

If we distribute and combine like terms, this is {eq}2x^2 + x - 2x - 1 - x^2 - x\\ x^2 - 2x - 1 {/eq}

and so the derivative of the function is given by {eq}\left(\dfrac{f(x)}{g(x)}\right)' = \dfrac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} = \dfrac{x^2 - 2x - 1}{x-1} {/eq}

The "-1" in the numerator of the derivative comes from distributing properly in the numerator of the quotient rule. The mistake in the work laid out in the question was stating that the derivative of {eq}x^2 + x {/eq} was {eq}2x {/eq} instead of {eq}2x+1 {/eq}