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Derivative of cubic absolute value function

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    f(x) = |x+2|^3 -1


    3. The attempt at a solution

    - So, I know the formula is: d|x| / dx = x / |x|

    - My guess would be: (x+2)^3 / |x + 2|^3 ???

    I'm trying to find Increase and Decrease intervals for the graph..
     
  2. jcsd
  3. Dec 7, 2011 #2

    eumyang

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    No. Looks like you'll need to use the chain rule.
     
  4. Dec 7, 2011 #3
    ok... hmmm , so maybe like this? :

    (3) |x+2|^2 ( x+2 / |x+2| )
     
  5. Dec 7, 2011 #4

    Char. Limit

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    That sounds right.
     
  6. Dec 7, 2011 #5
    Great, thanks... Now question #2:

    I'm testing for concavity, so I'm getting the 2nd derivative...

    For the x+2 / |x+2| part, does the quotient rule apply as normal??

    The procedure would be a chain rule involving a product rule and a quotient rule right?
     
  7. Dec 7, 2011 #6

    Char. Limit

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    It might help if you simplified a bit, getting rid of an |x+2| in the numerator and denominator. Then you just have 3 (x+2) |x+2|, and that's simple to solve with the product rule.
     
  8. Dec 7, 2011 #7
    You're right. So I guess the second derivative would be:

    f '' (x) = ( 3 |x+2| ) + 3 (x+2) (x+2) / |x+2|

    Thanks a lot btw
     
  9. Dec 7, 2011 #8

    SammyS

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    To find the derivative of [itex]\displaystyle \frac{x+2}{|x+2|}\,,[/itex] simply consider its graph. The derivative is undefined at x=-2 and is 1 everywhere else.
     
  10. Dec 7, 2011 #9

    SammyS

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    In my opinion, the clearest way to do this is with the piecewise definition of |x + 2|.

    [itex]\displaystyle f(x) = \left|x+2\right|^3=\left\{
    \matrix{(x+2)^3,\ \ \text{if}\quad x>-2 \\
    \ \ 0\ ,\ \ \quad\quad\quad\text{if}\quad x=-2 \\
    -(x+2)^3,\ \ \text{if}\quad x<-2} \right.[/itex]

    Finding the first & second derivatives is straight forward, if x ≠ -2. Take the limits from the left & right to show that f ' (-2) = 0 and f '' (-2) = 0
     
  11. Dec 7, 2011 #10

    Char. Limit

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    Considering that OP started with this as the expression for the derivative:

    [tex]f'(x)=3 |x+2|^2 \frac{x+2}{|x+2|}[/tex]

    It'd probably just be easier to cancel out an |x+2| term from the numerator and denominator. It's easy to find the second derivative then.
     
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