# Differentiate f(x) = \frac{e^{2x-3} + 5}{x}

## Question:

Differentiate {eq}\displaystyle f(x) = \frac{e^{2x-3} + 5}{x} {/eq}

## Quotient Rule:

We use the quotient rule if we want to derive rational functions or functions which are expressible as quotient of two functions.

For us to differentiate rational functions, we need to use the following formula

{eq}\displaystyle D_x \left(\frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2} {/eq}

Applying the quotient rule with {eq}f(x) = e^{2x-3} + 5 {/eq} and {eq}g(x) = x {/eq}:

{eq}\begin{align*} \displaystyle f(x) & = \frac{e^{2x-3} + 5}{x}\\ f'(x)& = \frac{ (x)D_x(e^{2x-3} + 5) -(e^{2x-3} + 5)D_x(x)}{x^2} \ \ \ \left[\displaystyle D_x \left(\frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}\right] \\ f'(x)& = \frac{ (x)(e^{2x-3} )(2)-(e^{2x-3} + 5)(1)}{x^2} \ \ \ \ \ \ \ \ \ \left[\mathrm{Chain \ Rule}: \ D_x(f(g(x)) = f'(g(x))g'(x)\right] \\ \implies f'(x)& = \frac { 2xe^{2x-3} -e^{2x-3} - 5}{x^2}\\ \end{align*} {/eq}