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

1 Answer
Ron-John Ahn
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)2(2x+2−x3−x2)9(x+1)4.

Explanation:

Using the quotient rule, you may let (x+1)^3(x+1)3=v and (x-1)=u.
In order to solve this you would use
(v*(du)/dx-u*(dv)/dx)/v^2v⋅dudx−u⋅dvdxv2

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

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

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

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