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

Jun 21, 2018

You've got to use both the quotient rule and the chain rule together, giving you $\frac{2 \left(2 x + 2 - {x}^{3} - {x}^{2}\right)}{9 {\left(x + 1\right)}^{4}}$.

#### Explanation:

Using the quotient rule, you may let ${\left(x + 1\right)}^{3}$=v and (x-1)=u.
In order to solve this you would use
$\frac{v \cdot \frac{\mathrm{du}}{\mathrm{dx}} - u \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

where $\frac{\mathrm{du}}{\mathrm{dx}}$ is the derivative of (x-1) which = 1.

Then, $\frac{\mathrm{dv}}{\mathrm{dx}}$ is the derivative of ${\left(x + 1\right)}^{3}$ which you can find using the chain rule where ${\left(x + 1\right)}^{3}$ is in the form ${\left(a x + b\right)}^{n}$.

Simply, the chain rule can be used so that $\frac{\mathrm{dy}}{\mathrm{dx}} \cdot f \left(x\right)$
(where $f \left(x\right)$ is some ${\left(a x + b\right)}^{n}$), = $n {\left(a x + b\right)}^{n - 1} \cdot \left(a\right)$

$\therefore \frac{d}{\mathrm{dx}} \cdot {\left(x + 1\right)}^{3} = 3 {\left(x + 1\right)}^{2} \cdot \left(1\right) = 3 {\left(x + 1\right)}^{2}$

which, after expanding = $3 {x}^{2} + 6 x + 3$

Now with the knowledge of the derivatives of u and v, you can differentiate the function using the quotient rule by substituting in the v and u values:

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

=$\frac{{x}^{3} + {x}^{2} + x + 1 - 3 {x}^{3} - 6 {x}^{2} - 3 x + 3 {x}^{2} + 6 x + 3}{9 {\left(x + 1\right)}^{4}}$

After simplifying, this gives:

$\frac{- 2 {x}^{3} - 2 {x}^{2} + 4 x + 4}{9 {\left(x + 1\right)}^{4}}$

=$\frac{2 \left(2 x + 2 - {x}^{3} - {x}^{2}\right)}{9 {\left(x + 1\right)}^{4}}$