# How do you differentiate [(x^3-3x^2+4)/(x^2)]?

Jul 29, 2015

$1 - \frac{8}{x} ^ 3$

#### Explanation:

Simplify the expression as $x - 3 + \frac{4}{x} ^ 2$

Now differentiate term by term to get 1-$\frac{8}{x} ^ 3$

Jul 29, 2015

You could also differentiate this expression by using the quotient rule.

#### Explanation:

Alternatively, you can use the quotient rule to differentiate this function.

The quotient rule tells you that a function that can be written as the quotient of two other functions

$y = f \frac{x}{g} \left(x\right)$,

can be differentiated by using this formula

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

$f \left(x\right) = {x}^{3} - 3 {x}^{2} + 4$

and

$g \left(x\right) = {x}^{2}$

The derivative of $y$ will thus be equal to

${y}^{'} = \frac{\frac{d}{\mathrm{dx}} \left({x}^{3} - 3 {x}^{2} + 4\right) \cdot {x}^{2} - \left({x}^{3} - 3 {x}^{2} + 4\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2}\right)}{{x}^{2}} ^ 2$

${y}^{'} = \frac{\left(3 {x}^{2} - 6 x\right) \cdot {x}^{2} - \left({x}^{3} - 3 {x}^{2} + 4\right) \cdot 2 x}{x} ^ 4$

${y}^{'} = \frac{3 {x}^{4} - 6 {x}^{3} - 2 {x}^{4} + 6 {x}^{3} - 8 x}{x} ^ 4$

This is equivalent to

${y}^{'} = \frac{{x}^{4} - \textcolor{red}{\cancel{\textcolor{b l a c k}{6 {x}^{3}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{6 {x}^{3}}}} - 8 x}{x} ^ 4$

${y}^{'} = \frac{{x}^{4} - 8 x}{x} ^ 4$

FInally, you can simplify this expression to get

${y}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{4}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{4}}}}} - \frac{8 \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{x}}}}{{x}^{\textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{4}}}}} = 1 - \frac{8}{x} ^ 3$

If you want, you can go on to factor this by using the difference of two cubes

color(blue)(a^3 - b^3 = (a-b) * (a^2 + ab + b^2)

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