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Derivative with Absolute Value

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Find f'(x), if

    f(x) = [x^2 * (3x + 2)^(1/3)] / [(2x - 3)3]

    Where the absolute value symbol surrounds the entire function.

    2. Relevant equations
    N/A


    3. The attempt at a solution
    My attempts don't account for the absolute value of the function. Otherwise, I can still take the derivative using quotient rule, product rule, and chain rule. Must I approach things differently because of the absolute value symbols? If so, how?
     
    Last edited: Oct 19, 2009
  2. jcsd
  3. Oct 19, 2009 #2

    Mark44

    Staff: Mentor

    Is this your function?
    [tex]\frac{|x^2(3x + 2)^{1/3}|}{(2x - 3)^3}[/tex]

    If so, you can simplify it a bit to this:
    [tex]\frac{x^2|(3x + 2)^{1/3}|}{(2x - 3)^3}[/tex]

    You will need to take into account the absolute values of the factor in the numerator and the one in the denominator. One way to do this is to look at your function on three different intervals: -infinity < x <-2/3, -2/3 < x < 3/2, and 3/2 < x < infinity.
    Each of these intervals gives a different version of your function if you remove the absolute values, so each will give you a different version of the derivative.
     
  4. Oct 19, 2009 #3

    \frac{x^2}{(-3+2 x)^3 (2+3 x)^{2/3}}-\frac{6 x^2 (2+3 x)^{1/3}}{(-3+2 x)^4}+\frac{2 x (2+3 x)^{1/3}}{(-3+2 x)^3}

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

     
  5. Oct 19, 2009 #4
    You can also use the chain rule knowing that |x| = √(x2)
     
  6. Oct 20, 2009 #5
    Ok; I think I understand how to approach this now. I'll get more help from my instructor to be sure. (I completed the assignment before, but wanted to understand for the final.)

    So, thanks!
     
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