# How do you use the quotient rule to differentiate f(x) = (x^1.7 + 2) / (x^2.8 + 1)?

Mar 6, 2016

$f ' \left(x\right) = \frac{1.7 {x}^{0.7} - 5.6 {x}^{1.8} - 1.1 {x}^{3.5}}{{x}^{2.8} + 1} ^ 2$

#### Explanation:

The quotient rule states that

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

In the given function, we see that

$g \left(x\right) = {x}^{1.7} + 2$
$h \left(x\right) = {x}^{2.8} + 1$

To find these functions' derivatives, use the power rule, which states that

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

Hence:

$g ' \left(x\right) = 1.7 {x}^{1.7 - 1} = 1.7 {x}^{0.7}$
$h ' \left(x\right) = 2.8 {x}^{2.8 - 1} = 2.8 {x}^{1.8}$

Plugging these functions into the quotient rule expression, we see that

$f ' \left(x\right) = \frac{\left({x}^{2.8} + 1\right) \left(1.7 {x}^{0.7}\right) - \left({x}^{1.7} + 2\right) \left(2.8 {x}^{1.8}\right)}{{x}^{2.8} + 1} ^ 2$

When simplifying, recall that ${x}^{a} \left({x}^{b}\right) = {x}^{a + b}$.

$f ' \left(x\right) = \frac{1.7 {x}^{3.5} + 1.7 {x}^{0.7} - 2.8 {x}^{3.5} - 5.6 {x}^{1.8}}{{x}^{2.8} + 1} ^ 2$

$f ' \left(x\right) = \frac{1.7 {x}^{0.7} - 5.6 {x}^{1.8} - 1.1 {x}^{3.5}}{{x}^{2.8} + 1} ^ 2$