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Math Help - Domain sizes

  1. #1
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    Domain sizes

    I came across this question in one of my old textbooks:

    If the function y = f(x) has domain D and its derivative y' = f'(x) has domain D', what can one say about the relative sizes of D and D'?

    After a bit of thought, I found 3 examples having D < D', D = D', and D > D':

    y = x^x \rightarrow y' = x^x (\ln x + 1)
    y = x^2 \rightarrow y' = 2x
    y = \ln x \rightarrow y' = \frac{1}{x}

    So, is the answer, it depends? Or am I missing something?

    -Scott
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  2. #2
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    Quote Originally Posted by ScottO View Post
    I came across this question in one of my old textbooks:

    If the function y = f(x) has domain D and its derivative y' = f'(x) has domain D', what can one say about the relative sizes of D and D'?

    After a bit of thought, I found 3 examples having D < D', D = D', and D > D':

    y = x^x \rightarrow y' = x^x (\ln x + 1)
    y = x^2 \rightarrow y' = 2x
    y = \ln x \rightarrow y' = \frac{1}{x}

    So, is the answer, it depends? Or am I missing something?

    -Scott
    Hello,

    1. You've found three different possible relations between the domain of a function and the domain of the first derivative of this function. But the first case you mention is impossible:

    If D(f)\subset D(f') then there exist at least one x for which f(x) doesn't exist but the derivative. Or in plain words: That can't be.
    The only relation between these two sets is: D(f) \subseteq D(f')

    2. Considering your examples:

    i) y = x^x ~,\ x\in\mathbb{R}_0^+ because the limit exist: \lim_{x\to 0+}\left(x^x\right)=1.
    But: y' = x^x (\ln x + 1)~,\ x\in\mathbb{R}^+ because ln(0) is not defined. So D(f')\subset D(f)

    ii) y = x^2 \rightarrow y' = 2x Obviously D(f) = D(f')

    iii) y = \ln x \rightarrow y' = \frac{1}{x} . The ln-function is defined for positive real numbers only and therefore the domain of the first derivation can't be greater than the domain of the function (see above). Thus D(f) = D(f')
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