# How do you find the derivative for [e^(1/2x)]/(2x^3)?

Jul 30, 2015

You can find this derivative by using the quotient rule and the chain rule.

#### Explanation:

Notice that your function can be written in the form

$y = {e}^{\frac{1}{2} x} / \left(2 {x}^{3}\right) = f \frac{x}{g} \left(x\right)$,

which means that you can use the quotient rule to find its derivative.

The derivative of a function that can be expressed as the quotient of two other functions can be found by using

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \frac{{f}^{'} \left(x\right) \cdot g \left(x\right) - f \left(x\right) \cdot {g}^{'} \left(x\right)}{g \left(x\right)} ^ 2}$, where $g \left(x\right) \ne 0$.

This means that you can write

${y}^{'} = \frac{\frac{d}{\mathrm{dx}} \left({e}^{\frac{1}{2} x}\right) \cdot \left(2 {x}^{3}\right) - {e}^{\frac{1}{2} x} \cdot \frac{d}{\mathrm{dx}} \left(2 {x}^{3}\right)}{2 {x}^{3}} ^ 2$

To find $\frac{d}{\mathrm{dx}} \left({e}^{\frac{1}{2} x}\right)$ you can use the chain rule, which helps you differentiate functions that depends on a variable $u$, which in turn depend on a variable $x$.

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(h \left(u\right)\right) = \frac{d}{\mathrm{du}} y \cdot \frac{d}{\mathrm{dx}} \left(u\right)}$

In your case, you have $h \left(u\right) = {e}^{u}$ and $u = \frac{1}{2} x$, so that

$\frac{d}{x} \left(h \left(u\right)\right) = \frac{d}{\mathrm{du}} {e}^{u} \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{2} x\right)$

$\frac{d}{\mathrm{dx}} \left(h \left(u\right)\right) = {e}^{u} \cdot \frac{1}{2} = \frac{1}{2} \cdot {e}^{\frac{1}{2} x}$

Plug this into your target derivative to get

${y}^{'} = \frac{\frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot {e}^{\frac{1}{2} x} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} {x}^{3} - {e}^{\frac{1}{2} x} \cdot 6 {x}^{2}}{4 {x}^{6}}$

${y}^{'} = \frac{{x}^{3} \cdot {e}^{\frac{1}{2} x} - 6 {x}^{2} \cdot {e}^{\frac{1}{2} x}}{4 {x}^{6}}$

This can be further simplified to get

y^' = (color(red)(cancel(color(black)(x^2))) * e^(1/2x) * (x - 6))/(4x^(color(red)(cancel(color(black)(6))) color(blue)(4))) = color(green)((e^(1/2x)(x-6))/(4x^4)

Alternatively, you could rewrite your original function as

$y = \frac{1}{2} \cdot {e}^{\frac{1}{2} x} \cdot {x}^{- 3}$

and use the product, chain, and power rules to get

${y}^{'} = \frac{1}{2} \cdot \left[\frac{d}{\mathrm{dx}} \left({e}^{\frac{1}{2} x}\right) \cdot {x}^{_ 3} + {e}^{\frac{1}{2} x} \cdot \frac{d}{\mathrm{dx}} \left({x}^{- 3}\right)\right]$

${y}^{'} = \frac{1}{2} \cdot \left[\frac{1}{2} \cdot {e}^{\frac{1}{2} x} \cdot {x}^{- 3} + {e}^{\frac{1}{2} x} \cdot \left(- 3\right) {x}^{- 4}\right]$

${y}^{'} = \frac{1}{2} \cdot {e}^{\frac{1}{2} x} \left(\frac{1}{2 {x}^{3}} - \frac{3}{x} ^ 4\right)$

which will get you

${y}^{'} = \frac{1}{2} \cdot {e}^{\frac{1}{2} x} \cdot \frac{\left(x - 6\right)}{2 {x}^{4}} = \textcolor{g r e e n}{\frac{{e}^{\frac{1}{2} x} \cdot \left(x - 6\right)}{4 {x}^{4}}}$