# How do you differentiate f(x) =(1-x)/(x^3-6) using the quotient rule?

Sep 17, 2017

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

#### Explanation:

Let's start off with the definition of the quotient rule:

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

Or to make remembering the rule simpler, let $f \left(x\right)$ be "$h i g h$," being the upper function and let $g \left(x\right)$ be "$l o w$," being the lower function to give us:

$\frac{d}{\mathrm{dx}} \left(\frac{h i g h}{l o w}\right) = \frac{l o w \cdot d ' \left(h i g h\right) - \left(h i g h \cdot d ' \left(l o w\right)\right)}{l o w} ^ 2$

where $d '$ denotes the "derivative of."

This formula can be read as "low d high minus high d low over low squared," which has a nice flow for memorization.

In the case $f \left(x\right) = \frac{1 - x}{{x}^{3} - 6}$, let "$l o w$" denote ${x}^{3} - 6$ and "$h i g h$" denote $1 - x$

Applying the quotient rule,

$\frac{d}{\mathrm{dx}} \left(\frac{h i g h}{l o w}\right) = \frac{\left({x}^{3} - 6\right) \cdot d ' \left(1 - x\right) - \left(\left(1 - x\right) \cdot d ' \left({x}^{3} - 6\right)\right)}{{x}^{3} - 6} ^ 2$

Now switching $\left(\frac{h i g h}{l o w}\right)$ to $f \left(x\right)$ for simplicity,

d/dx (f(x)) = ((x^3 - 6)⋅(-1) - ((1-x)*(3x^2)))/(x^3 - 6)^2

Knowing that by the power rule $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$

And that the derivative of a constant is $0$.

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

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