# How do you find the derivative of f(x)=x^3/(x^3-5x+10)?

Mar 25, 2018

$= - 10 \frac{{x}^{2} \left(x - 3\right)}{{x}^{3} - 5 x + 10} ^ 2$

#### Explanation:

Use the quotient rule:

Let $u = f \left(x\right)$ and $v = g \left(x\right)$

Then $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$

For this question:

$u = {x}^{3}$
$u ' = 3 {x}^{2}$

$v = {x}^{3} - 5 x + 10$
$v ' = 3 {x}^{2} - 5$

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{3 {x}^{2} \left({x}^{3} - 5 x + 10\right) - {x}^{3} \left(3 {x}^{2} - 5\right)}{{x}^{3} - 5 x + 10} ^ 2$

$= \frac{\cancel{3 {x}^{5}} - 15 {x}^{3} + 30 {x}^{2} \cancel{- 3 {x}^{5}} + 5 {x}^{3}}{{x}^{3} - 5 x + 10} ^ 2$

$= \frac{- 10 {x}^{3} + 30 {x}^{2}}{{x}^{3} - 5 x + 10} ^ 2$

$= - 10 \frac{{x}^{2} \left(x - 3\right)}{{x}^{3} - 5 x + 10} ^ 2$