# Using the quotient rule to differentiate the function, f(x) = x^2/\tan x - x.

## Question:

Using the quotient rule to differentiate the function{eq}, f(x) = \dfrac {x^2}{\tan x - x}. {/eq}

## Quotient rule of differentiation:

If {eq}s {/eq} and {eq}t {/eq} are differentiable, then {eq}r(x) = \dfrac {s(x)}{t(x)} {/eq} is also differentiable,

then the derivative of the function with respect to {eq}x {/eq} is given by,

{eq}r'(x) = \dfrac {s'(x) \cdot t(x) - t'(x) \cdot s(x)}{(t(x))^2} {/eq}

Recall that the derivative of {eq}x^n {/eq} is {eq}nx^{n - 1} {/eq}, where {eq}n {/eq} is a real number.

Differentiate the given function using the quotient rule,

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