Question

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

Answer 1

You've got to use both the quotient rule and the chain rule together, giving you `(2(2x+2-x^3-x^2))/(9(x+1)^4)`.

Explanation:

Using the quotient rule, you may let `(x+1)^3`=v and (x-1)=u.
In order to solve this you would use
`(v*(du)/dx-u*(dv)/dx)/v^2`

where `(du)/dx` is the derivative of (x-1) which = 1.

Then, `(dv)/dx` is the derivative of `(x+1)^3` which you can find using the chain rule where `(x+1)^3` is in the form `(ax+b)^n`.

Simply, the chain rule can be used so that `(dy)/dx*f(x)`
(where `f(x)` is some `(ax+b)^n`), = `n(ax+b)^(n-1)*(a)`

`therefore d/dx*(x+1)^3=3(x+1)^2*(1)=3(x+1)^2`

which, after expanding = `3x^2+6x+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:

`((x+1)^3*(1)-(x-1)*(3x^2+6x+3))/(3(x+1)^2)^2`

=`(x^3+x^2+x+1-3x^3-6x^2-3x+3x^2+6x+3)/(9(x+1)^4)`

After simplifying, this gives:

`(-2x^3-2x^2+4x+4)/(9(x+1)^4)`

=`(2(2x+2-x^3-x^2))/(9(x+1)^4)`

Which is your final answer.