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Domain of a derivative

  1. Jul 12, 2006 #1
    Can the domain of f'(x) be larger than the domain of f(x)?

    Here's my work, as per the rules:smile:

    The problem I'm working on states
    f(x) = x + √x
    and says to find f'(x), the domain of f(x), and the domain of f'(x).

    I found that f'(x)=1, so I want to say that its domain is all real numbers. But the domain of f(x) is only (0, ∞).

    Hence my question in the first line.
  2. jcsd
  3. Jul 12, 2006 #2
    Look at your differentiation again. You only calculated the derivative of f(x) = x.
  4. Jul 12, 2006 #3
    You need to differentiate Sqrt(x) too, as mentionned by Hammie and then see for what values of x D[f(x) = x + Sqrt(x)] is defined.

    By that I mean check for what values of x the denominator of D[f(x)] does not equal zero. If the denominator of D[f(x)] equals zero, then D[f(x)] is not defined for these particular values of x because dividing by zero is one of the few "forbidden" operations in math.

    To answer your question, the domain of D[f(x)] can include the domain of f(x). For example f(x) = ln(x)
    The domain is ]0, +infinity[ and D[f(x)] = 1/x had R/{0} as domain.

    Hope it helps
  5. Jul 12, 2006 #4


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    Staff Emeritus
    Science Advisor

    is defined as
    [tex]lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}[/tex]
    If f(a) is not defined, then f '(a) is not defined either.
  6. Jul 14, 2006 #5
    I re-did the problem and came up with f'(x) = 1 + [1/(2√ x)].
    Is that the right answer? (It's an even problem, so the book doesn't have the answer...)
    Thank you!
  7. Jul 14, 2006 #6
    Looks good to me. So, what does that say about the domain?
    Last edited: Jul 14, 2006
  8. Jul 14, 2006 #7
    That means that my original question is irrelevant for this problem, b/c the domain of f'(x) is x=0 or x>0, which the domain for f'(x).

    Thanks everyone, both for checking my work and answering my original question!
  9. Jul 14, 2006 #8


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    Staff Emeritus
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    No, it's not.
    [tex]f(x)= x+ \sqrt{x}[/tex]
    is defined for x= 0 but
    [tex]f '(x)= 1- \frac{1}{2\sqrt{x}}[/tex]
    is not!

    And the answer to your original question is "No, the domain of f ' cannot be larger than the domain of f, but it can be a proper subset."
  10. Jul 15, 2006 #9
    Thank you a lot HallsofIvy for correcting my mistakes. I had mistaken the derivative ( the linear application wich associates the function to it's linear approx. ) and the function it represents.
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