# How do you find the derivative of F(x)= (2x^3-1)/x^2?

Dec 9, 2017

The quotient rule states that given a function $F \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} , \frac{\mathrm{dF}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \cdot h \left(x\right) - g \left(x\right) \cdot \frac{\mathrm{dh}}{\mathrm{dx}}}{h \left(x\right)} ^ 2$. See explanation for solution.

#### Explanation:

The quotient rule states that given a function $F \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} , \frac{\mathrm{dF}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \cdot h \left(x\right) - g \left(x\right) \cdot \frac{\mathrm{dh}}{\mathrm{dx}}}{h \left(x\right)} ^ 2$

The power rule can help us find the derivatives of both numerator and denominator quickly.

$g \left(x\right) = 2 {x}^{3} - 1 , \frac{\mathrm{dg}}{\mathrm{dx}} = 6 {x}^{2}$
$h \left(x\right) = {x}^{2} , \frac{\mathrm{dh}}{\mathrm{dx}} = 2 x , {h}^{2} \left(x\right) = {x}^{4}$

Thus...

$\frac{\mathrm{dF}}{\mathrm{dx}} = \frac{\left(6 {x}^{2}\right) \left({x}^{2}\right) - \left(2 {x}^{3} - 1\right) \left(2 x\right)}{x} ^ 4 = \frac{6 {x}^{4} - 4 {x}^{4} + 2 x}{{x}^{4}} = \frac{2 {x}^{4} + 2 x}{x} ^ 4 = 2 + \frac{2}{x} ^ 3$

Of note, the function will approach $- \infty$ as $x \to 0$

Dec 10, 2017

I would rewrite: $F \left(x\right) = 2 x - \frac{1}{x} ^ 2$

#### Explanation:

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

using $\frac{d}{\mathrm{dx}} \left(- 1 {x}^{-} 2\right) = 2 {x}^{-} 3$

and rewriting as a single quotient.