# How do you find the derivative for (x^2 + 3) / x?

Jul 31, 2015

I found: $\frac{{x}^{2} - 3}{x} ^ 2$

#### Explanation:

You can use the Quotient Rule where the derivative of $f \frac{x}{g} \left(x\right)$ is:
$\frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

$\frac{2 x \cdot x - \left({x}^{2} + 3\right) \cdot 1}{x} ^ 2 = \frac{2 {x}^{2} - {x}^{2} - 3}{x} ^ 2 = \frac{{x}^{2} - 3}{x} ^ 2$

Jul 31, 2015

Alternatively, you could differentiate this function by using the product rule.

#### Explanation:

You could rewrite your original function as

y = (x^2 + 3) * x^(-1

This will allow you to use the product rule, which tells you that functions that can be expressed as the product of two other functions

$y = f \left(x\right) \cdot g \left(x\right)$

can be differentiate by

color(blue)(d/dx(y) = f^'(x)g(x) + f(x)g^'(x)

$f \left(x\right) = {x}^{2} + 3$ and $g \left(x\right) = {x}^{- 1}$, so that your derivative will be

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

$\frac{d}{\mathrm{dx}} \left(y\right) = 2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}} + \left({x}^{2} + 3\right) \cdot \left(- {x}^{- 2}\right)$

$\frac{d}{\mathrm{dx}} \left(y\right) = 2 - \frac{{x}^{2} + 3}{x} ^ 2$

$\frac{d}{\mathrm{dx}} \left(y\right) = 2 - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}}} - \frac{3}{x} ^ 2 = \textcolor{g r e e n}{1 - \frac{3}{x} ^ 2}$

Jul 31, 2015

Alternatively again, you could use the power rule.

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} + 3}{x}$

$= {x}^{2} / x + \frac{3}{x}$

$= x + \frac{3}{x}$

$= x + 3 {x}^{- 1}$

$f ' \left(x\right) = 1 - 3 {x}^{- 2}$

$f ' \left(x\right) = 1 - \frac{3}{x} ^ \left(2\right)$