# Differentiate the given function with respect to x. f(x) = {x^3 - 2} / {x - 9}

## Question:

Differentiate the given function with respect to {eq}x {/eq}.

$$f(x) = \frac{x^3 - 2}{x - 9}$$

## Quotient rule of differentiation:

We will use the quotient rule to differentiate the function.

The quotient rule states that the derivative of {eq}\dfrac uv {/eq} is {eq}\displaystyle \frac {u'v - v'u}{v^2}, \ \ v \neq 0 {/eq}.

We have,

$$f(x) = \frac{x^3 - 2}{x - 9}$$

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

\displaystyle \begin{align*} f'(x) &= \frac {d}{dx} \left [ \frac{x^3 - 2}{x - 9} \right ] \\[0.3 cm] &= \frac{\frac {d}{dx}\left [ x^3 - 2 \right ] \cdot (x - 9) - \frac {d}{dx}\left [ x - 9 \right ] \cdot \left ( x^3 - 2 \right )}{(x - 9)^2} \\[0.3 cm] &= \frac{\left [ 3x^2 - 0 \right ] \cdot (x - 9) - \left [ 1 - 0 \right ] \cdot \left ( x^3 - 2 \right )}{(x - 9)^2} \\[0.3 cm] &= \frac{3x^2 (x - 9) - \left ( x^3 - 2 \right )}{(x - 9)^2} \\[0.3 cm] &= \frac{3x^3 - 27x^2 - x^3 + 2 }{(x - 9)^2} \\[0.3 cm] f'(x) &= \frac{2x^3 - 27x^2 + 2 }{(x - 9)^2} \\[0.3 cm] \end{align*} 