# How do you find the derivative of h(s)=((3s^2)-s+1)/s^2?

Aug 21, 2015

${h}^{'} = \frac{s - 2}{s} ^ 3$

#### Explanation:

You can differentiate this function by using the quotient rule, which allows you to differentiate functions that take the form

$\textcolor{b l u e}{h \left(x\right) = f \frac{x}{g} \left(x\right)} \text{ }$, where $\textcolor{b l u e}{g \left(x\right) \ne 0}$

by using the formula

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(h \left(x\right)\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) - f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}{g \left(x\right)} ^ 2}$

$f \left(s\right) = 3 {s}^{2} - 2 + 1 \text{ }$ and $\text{ } g \left(s\right) = {s}^{2}$

This means that you can write

$\frac{d}{\mathrm{ds}} \left(h \left(s\right)\right) = \frac{\left[\frac{d}{\mathrm{ds}} \left(3 {s}^{2} - s + 1\right) \cdot {s}^{2} - \left(3 {s}^{2} - s + 1\right) \cdot \frac{d}{\mathrm{dx}} \left({s}^{2}\right)\right)}{{s}^{2}} ^ 2$

${h}^{'} = \frac{\left(6 s - 1\right) \cdot {s}^{2} - \left(3 {s}^{2} - s + 1\right) \cdot 2 s}{s} ^ 4$

${h}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6 {s}^{3}}}} - {s}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{6 {s}^{3}}}} + 2 {s}^{2} - 2 s}{s} ^ 4$

${h}^{'} = \frac{{s}^{2} - 2 s}{s} ^ 4$

You can simplify this further to get

${h}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{s}}} \cdot \left(s - 2\right)}{s} ^ \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} = \textcolor{g r e e n}{\frac{s - 2}{s} ^ 3}$