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

Question

Christopher Agresta · Accepted Answer

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)#.

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.

Evelyn Agard · Answer

undefined To differentiate $ f(x) = \frac{{x - 1}}{{(x + 1)^3}} $ using the quotient rule, follow these steps:

1. Identify $ u(x) $ as the numerator and $ v(x) $ as the denominator.
2. Apply the quotient rule: $ f'(x) = \frac{{u'(x)v(x) - u(x)v'(x)}}{{[v(x)]^2}} $.
3. Find $ u'(x) $ and $ v'(x) $.
4. Substitute the values into the quotient rule formula.
5. Simplify the expression.