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Is the function differentiable at x = p?

  1. Jan 8, 2016 #1
    Hello mates,

    is the function ## f(x) = \frac{(2^x - 1)}{x} ## differentiable at x = 0? For it to be differentiable it has to be continuous? From the graph f(0) is undefined although limit exists. I have read that at points like a corner, gap and vertical tangents it is not differentiable. So why is it differentiable at f(0)?

    Sehr Danke!
     
  2. jcsd
  3. Jan 8, 2016 #2
    Quotient rule:
    if ##f(x) = g(x) / h(x)##, ##g(x) = 2^x - 1## and ##h(x) = x## then ##f'(x) = (h(x) * g'(x) - g(x) * h'(x) / g(x)^2##. In this case, ##g(x)^2 = 0## so the derivative would be undefined. Does this help?
     
  4. Jan 8, 2016 #3

    Mark44

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    No.
    This function isn't defined at x = 0, hence it's not continuous at x = 0, so it can't be differentiable there.
     
  5. Jan 8, 2016 #4

    SammyS

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    However, ##\displaystyle \ \lim _{x\to0} f'(x)\ ## does exist.
     
  6. Jan 8, 2016 #5

    WWGD

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    This assumes ## g \neq 0 ## which is not the case, like many said. You may say that _at points where ##g \neq 0 ##_ (since both ## f,g## are everywhere differentiable), you can say ## (f/g)' = \frac {f'g-fg'}{g^2}= \frac {(ln2)(2^x)x -2^x}{x^2} ##, using the identity ## 2^x = e^{xln2}##.
     
  7. Jan 9, 2016 #6

    pasmith

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    Functions can only be differentiable where they are actually defined. However the singularity at the origin is removable by setting [itex]f(0) = \ln 2[/itex]. Then the function is not merely differentiable at the origin but analytic, and its Taylor series is [tex]
    f(x) = \sum_{n=0}^\infty \frac{(\ln 2)^{n+1} x^n}{(n+1)!}.[/tex]
     
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