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Question about domain of derivatives

  1. Aug 20, 2010 #1
    I was thinking how do I differentiate the domain of functions...

    Suppose I have a function:

    [tex]f(x) = \left\{\begin{matrix}
    x^2 -1, \;\; |x| \leq 1\\
    1 - x^2, \;\; |x| > 1

    \end{matrix}\right.[/tex]

    And I need to derive it:

    [tex]f'(x) = \left\{\begin{matrix}
    2x, \;\; |x| \leq 1\\
    -2x, \;\; |x| > 1

    \end{matrix}\right.[/tex]

    1. What do I do with the conditions of |x| ? Is there a rule to handle this?
    2. What will be the domain of f'(x)? Will the domain of f'(x) be the domain of f(x) minus the points where the function is non-differentiable?

    Thank you,
    Rafael Andreatta
     
  2. jcsd
  3. Aug 20, 2010 #2
    With regards to the derivative function for single variable functions, the maximal domain of the function that gives the derivative of f at each point will be the set of points in the domain of f where f is differentiable.
    For multivariable and vector-valued functions, a more sophisticated definition of the derivative at a point and the derivative function are necessary.
    The derivative at a point is generalized to a linear function over the tangent space to the point in the image of the function, and the derivative function is then a map from points in the domain of f to the space of linear functions over the tangent space to each point.
    This simplifies to the single-variable case, as linear functions over the tangent line to the curve are constants. From this perspective, the domain would be isomorphic, but not strictly the same.
     
    Last edited: Aug 20, 2010
  4. Aug 20, 2010 #3

    statdad

    User Avatar
    Homework Helper

    In a problem like this one (finding the derivative of a piecewise-defined function) I would state the work steps this way (this is my explanation only, not written to be textbook perfect)

    * Use ordinary derivative rules everywhere except for the domain values where the different pieces are put together
    * Examine the x-values where the pieces are joined. If the one-sided derivatives there agree, no problem. If the one-sided derivatives do not agree, your original function is not differentiable there.

    Note: my original first example had an error, pointed out by l'Hopital below: read my later post for the correction.

    However, if I have

    [tex]
    h(x) =
    \begin{cases}
    x^2 + 3x, \quad x \le 2 \\
    5x^2 - 5x, \quad x > 2
    \end{cases}
    [/tex]

    I need to examine the two one-sided derivatives at x = 2. The left-hand derivative there
    is [itex] 7 [/itex], while the right-hand derivative is [itex] 15 [/itex]. Since the left- and right-hand derivatives don't agree there, h is not differentiable there . Thus, for my second example (notice that [tex] \le 2 [/tex] has been replaced by [tex] < 2[/tex])

    [tex]
    h'(x) =
    \begin{cases}
    2x + 3, \quad x < 2 \\
    10x - 5, \quad x > 2
    \end{cases}
    [/tex]

    Hope this helps.
     
    Last edited: Aug 20, 2010
  5. Aug 20, 2010 #4
    I have a question regarding g in statdad's post. It is clear g is not continuous at x = 0. Does it really make sense for g to have a derivative x = 0 ?
     
  6. Aug 20, 2010 #5

    statdad

    User Avatar
    Homework Helper

    Well poop (that's not what I said to myself when I saw your email pop up, but you should get the drift)

    Perfect point and I have no excuse for my screwup. Let me use this example.

    [tex]
    f(x) = \begin{cases}
    x^2 + 6x - 7, \quad x \le 1 \\
    -9 + 10x - x^2, \quad x > 1
    \end{cases}
    [/tex]

    A quick check now, using the ideas in my earlier post, gives

    [tex]
    f'(x) = \begin{cases}
    2x + 6, \quad x \le 1\\
    10-2x, \quad x > 1
    \end{cases}
    [/tex]

    The idea of my second example still works. Huge apologies for any confusion this may have caused, and a big "Thank-you" to l'Hopital for being a fantastic proof reader.
     
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