# How do you use the quotient rule to differentiate y=(2x^4-3x)/(4x-1)?

Jul 19, 2018

$f ' \left(x\right) = \frac{24 {x}^{4} - 8 {x}^{3} + 3}{4 x - 1} ^ 2$

#### Explanation:

The Quotient rle is given by
$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$
let us denote by
$u = 2 {x}^{4} - 3 x$

and

$v = 4 x - 1$

so we get
$u ' = 8 {x}^{3} - 3$
and
$v ' = 4$

now we Can build the derivative:
$f ' \left(x\right) = \frac{\left(8 {x}^{3} - 3\right) \left(4 x - 1\right) - \left(2 {x}^{4} - 3 x\right) \cdot 4}{4 x - 1} ^ 2$

multiplying out the numerator and collecting like Terms we get

$f ' \left(x\right) = \frac{24 {x}^{4} - 8 {x}^{3} + 3}{4 x - 1} ^ 2$

Jul 19, 2018

$\frac{24 {x}^{4} - 8 {x}^{3} + 4}{4 x - 1} ^ 2$

#### Explanation:

If we have a quotient of functions $f \left(x\right)$ and $g \left(x\right)$, we can find the derivative with the Quotient Rule

$\frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

In our example, we essentially have

$f \left(x\right) = 2 {x}^{4} - 3 x \implies f ' \left(x\right) = 8 {x}^{3} - 3$ and

$g \left(x\right) = 4 x - 1 \implies g ' \left(x\right) = 4$

We can plug in our expressions into the Quotient Rule to get

$\frac{\left(8 {x}^{3} - 3\right) \left(4 x - 1\right) - 4 \left(2 {x}^{4} - 3 x\right)}{4 x - 1} ^ 2$

With FOIL and some algebraic distribution, we can simplify this expression to get

$\frac{24 {x}^{4} - 8 {x}^{3} + 4}{4 x - 1} ^ 2$

Hope this helps!