# Differentiate: f(x) = x^2 + 1 / x - 1

## Question:

Differentiate: {eq}f(x) = \dfrac{x^2 + 1}{x - 1} {/eq}

## Quotient rule of differentiation:

The quotient rule will allow us find the derivative of a function that has two function in ratio,

i.e., If {eq}u {/eq} and {eq}v {/eq} are two differentiable function, then the function {eq}\left ( \dfrac uv \right ) {/eq} is also differentiable,

then by quotient rule its derivative is given by,

{eq}\left ( \dfrac uv \right )' = \dfrac {u'v - v'u}{v^2} {/eq}

We have,

$$f(x) = \dfrac{x^2 + 1}{x - 1}$$

Differentiating the function with respect to {eq}x {/eq},

\displaystyle \begin{align*} f'(x) &= \dfrac {d}{dx}\left ( \dfrac{x^2 + 1}{x - 1} \right ) \\ &= \dfrac{\dfrac {d}{dx}\left ( x^2 + 1 \right ) \cdot (x - 1) - \dfrac {d}{dx} (x - 1) \cdot \left ( x^2 + 1 \right )}{(x - 1)^2} \\ &= \dfrac{\left ( 2x + 0 \right ) \cdot (x - 1) - (1 - 0) \cdot \left ( x^2 + 1 \right )}{(x - 1)^2} \\ &= \dfrac{2x (x - 1) - \left ( x^2 + 1 \right )}{(x - 1)^2} \\ &= \dfrac{2x^2 - 2x - x^2 - 1}{(x - 1)^2} \\ f'(x) &= \dfrac{x^2 - 2x - 1}{(x - 1)^2} \\ \end{align*} 