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

Feb 19, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 {x}^{3} + 3 {x}^{2} + 1}{4 x + 1} ^ 2$

#### Explanation:

The quotient rule states that

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

When we apply this to the given function $y$, we see that

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

Differentiating each of these through the power rule gives:

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

From here, distribute and combine like terms.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12 {x}^{3} + 4 x + 3 {x}^{2} + 1 - 4 {x}^{3} - 4 x}{4 x + 1} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 {x}^{3} + 3 {x}^{2} + 1}{4 x + 1} ^ 2$