How do you find the derivative of [(2x^3) - (4x^2) + 3] / x^2 ?

Mar 4, 2018

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

Explanation:

We use the quotient rule, which says for $f \left(x\right) = g \frac{x}{h \left(x\right)} , f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - h ' \left(x\right) g \left(x\right)}{{\left(h \left(x\right)\right)}^{2}}$.

So, we first let $f \left(x\right) = \frac{2 {x}^{3} - 4 {x}^{2} + 3}{x} ^ 2$.

Thus, $g \left(x\right) = 2 {x}^{3} - 4 {x}^{2} + 3$ and $h \left(x\right) = {x}^{2}$.

Next, we find $g ' \left(x\right)$ and $h ' \left(x\right)$

This just involves the power rule, so
$g ' \left(x\right) = 3 \cdot 2 {x}^{2} - 2 \cdot 4 x + 0 = 6 {x}^{2} - 8 x$
$h ' \left(x\right) = 2 x$

Now, we just plug these values into the quotient rule to get the following:
$f ' \left(x\right) = \frac{\left({x}^{2}\right) \left(6 {x}^{2} - 8 x\right) - \left(2 x\right) \left(2 {x}^{3} - 4 {x}^{2} + 3\right)}{{\left({x}^{2}\right)}^{2}}$
$f ' \left(x\right) = \frac{\left(6 {x}^{4} - 8 {x}^{3}\right) - \left(4 {x}^{4} - 8 {x}^{3} + 6 x\right)}{{x}^{4}}$
$f ' \left(x\right) = \frac{6 {x}^{4} - 4 {x}^{4} - 8 {x}^{3} + 8 {x}^{3} - 6 x}{{x}^{4}}$
$f ' \left(x\right) = \frac{2 {x}^{4} - 6 x}{{x}^{4}}$
$f ' \left(x\right) = \frac{2 x \left({x}^{3} - 3\right)}{{x}^{4}}$
$f ' \left(x\right) = \frac{2 \left({x}^{3} - 3\right)}{x} ^ 3$
Therefore, $\frac{d}{\mathrm{dx}} \left(\frac{2 {x}^{3} - 4 {x}^{2} + 3}{x} ^ 2\right) = \frac{2 \left({x}^{3} - 3\right)}{x} ^ 3$