# How do you differentiate f(x) = (x)/(x^4-x^3+6) using the quotient rule?

Feb 7, 2016

$f ' \left(x\right) = \frac{- 3 {x}^{4} + 2 {x}^{3} + 6}{{x}^{4} - {x}^{3} + 6} ^ 2$

#### Explanation:

The quotient rule states that

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

Applying this to the given function, we see that

$f ' \left(x\right) = \frac{\left({x}^{4} - {x}^{3} + 6\right) \frac{d}{\mathrm{dx}} \left[x\right] - x \frac{d}{\mathrm{dx}} \left[{x}^{4} - {x}^{3} + 6\right]}{{x}^{4} - {x}^{3} + 6} ^ 2$

Both of these derivatives can be found through the power rule:

$f ' \left(x\right) = \frac{\left({x}^{4} - {x}^{3} + 6\right) \left(1\right) - \left(x\right) \left(4 {x}^{3} - 3 {x}^{2}\right)}{{x}^{4} - {x}^{3} + 6} ^ 2$

Simplified, this gives

$f ' \left(x\right) = \frac{{x}^{4} - {x}^{3} + 6 - 4 {x}^{4} + 3 {x}^{3}}{{x}^{4} - {x}^{3} + 6} ^ 2$

$f ' \left(x\right) = \frac{- 3 {x}^{4} + 2 {x}^{3} + 6}{{x}^{4} - {x}^{3} + 6} ^ 2$