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

1 Answer
Jun 21, 2018

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.