# Do you use the quotient rule in the problem (x/(1-x))^3?

Mar 29, 2015

Yes, but it's not as simple as saying yes:

Written: $f \left(x\right) = {\left(\frac{x}{1 - x}\right)}^{3}$, this function is the cube of a simpler function. We'll need the power rule and the chain rule.

The derivative of a cube is: $\frac{d}{\mathrm{dx}} \left({u}^{3}\right) = 3 {u}^{2} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

In this problem, $u = \frac{x}{1 - x}$, so you'll need the quotient rule to find $\frac{\mathrm{du}}{\mathrm{dx}}$

$f ' \left(x\right) = 3 {\left(\frac{x}{1 - x}\right)}^{2} \left[\frac{1 \left(1 - x\right) - x \left(- 1\right)}{1 - x} ^ 2\right]$

$= 3 {\left(\frac{x}{1 - x}\right)}^{2} \left[\frac{1}{1 - x} ^ 2\right] = \frac{3 {x}^{2}}{1 - x} ^ 4$

If you rewrite the function as: $f \left(x\right) = {x}^{3} / {\left(1 - x\right)}^{3}$ You'll start with the quotient rule, and you'll need the power rule and chain rule when you differentiate the denominator.

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

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

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