Expert: Paul Klarreich - 2/13/2008

Question
This is on the Marginal Analysis in Business and Economics section.  The question is to use the general power rule and then other rules to find the derivative.  
f(x)=cube root over all x-x^2over x^3-4

Answer
Questioner:   Christine
Category:  Calculus
Private:  No
 
Subject:  Calculus
Question:  This is on the Marginal Analysis in Business and Economics section.  The question is to use the general power rule and then other rules to find the derivative.  
f(x)=cube root over all x-x^2over x^3-4
==============================================
Hi, Christine,

first, you have to know what the G.P.R. says:

It is written:

D[u^n] = n u^(n-1) du/dx.

So you have to know what those symbols say:

u is a function of x.  du/dx is the derivative of that function.

So it says to multiply two things.

Now to your example:

f(x) = cuberoot( (x - x^2)/(x^3 - 4) )

Basic fact about roots:  They can be written as powers using a fractional exponent.

f(x) = ( (x - x^2)/(x^3 - 4) )^(1/3)

Now it's just pattern-matching:

u = (x - x^2)/(x^3 - 4)

n = 1/3, and it matches  u^n.

Now we have to do some work.

Easy part:  nu^(n-1) = (1/3) u ^(1/3 - 1) =

nu^(n-1) = (1/3) u ^(-2/3)

= (1/3) ((x - x^2)/(x^3 - 4)) ^(-2/3)  << FIRST PART

Now to get du/dx.

u is a quotient.  So you use the quotient rule:

du   ()() - ()()
-- = -----------
dx       ()^2

du   (x^3 - 4)(1 - 2x) - (x - x^2)(3x^2)
-- = ------------------------------------
dx       (x^3 - 4)^2

Simplify a bit: [Yeah, right.  A bit, he says.]

du   x^3 - 4 - 2x^4 + 8x  - 3x^3 + 3x^4
-- = ------------------------------------
dx       (x^3 - 4)^2


du    - 4 + 8x  - 2x^3 + x^4
-- = -------------------------  << SECOND PART
dx       (x^3 - 4)^2

OK, put it together:

                                           - 4 + 8x  - 2x^3 + x^4
f'(x) = (1/3) ((x - x^2)/(x^3 - 4)) ^(-2/3) -------------------------
                                                (x^3 - 4)^2

You were expecting something simple?