# How do you find the derivative of y=(x^3+x)/(4x+1)?

Dec 25, 2015

$y = \frac{8 {x}^{3} + 3 {x}^{2} + 1}{4 x + 1} ^ 2$

#### Explanation:

Use the quotient rule:

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

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

So,

$f \left(x\right) = {x}^{3} + x$
$g \left(x\right) = 4 x + 1$

$f ' \left(x\right) = 3 {x}^{2} + 1$
$g ' \left(x\right) = 4$

Plug these in:

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

$\implies \frac{12 {x}^{3} + 3 {x}^{2} + 4 x + 1 - 4 {x}^{3} - 4 x}{4 x + 1} ^ 2$

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