# Differentiate the given function: f(x)=\frac{x}{(2x-5)^2}

## Question:

Differentiate the given function: {eq}f(x)=\frac{x}{(2x-5)^2} {/eq}

## Quotient Rule of Differentiation:

We are given a function defined by the ratio between two functions

{eq}\displaystyle f(x) = \frac{g(x) } { h(x) } {/eq}

The derivative of the function f found by means of the Quotient Rule of Differentiation:

{eq}\displaystyle f'(x) = \frac{g'(x)h(x) - g(x)h'(x) } { h^2(x) }. {/eq}

## Answer and Explanation:

The derivative of the funciton

{eq}\displaystyle f(x)=\frac{x}{(2x-5)^2} {/eq}

is calculated by means of the Quotient Rule of Differentiation

{eq}\displaystyle (f/g)' = \frac{g'h - gh' } { h^2 }. {/eq}

Therefore we get

{eq}\displaystyle f'(x)=\frac{\frac{d}{dx}\left(x\right)\left(2x-5\right)^2-\frac{d}{dx}\left(\left(2x-5\right)^2\right)x}{\left(\left(2x-5\right)^2\right)^2} \\ \displaystyle = \frac{1\cdot \left(2x-5\right)^2-4\left(2x-5\right)x}{\left(\left(2x-5\right)^2\right)^2} \\ \displaystyle = \frac{-5-2x}{\left(2x-5\right)^3}. {/eq}