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Derivative question

  1. Jul 30, 2007 #1

    daniel_i_l

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    1. The problem statement, all variables and given/known data
    This isn't exactly a homework question, it's more of a general one. Let's say for example that I have the function f(x) = arcsin((2x/(1+x^2)). I know that arcsin has no derivative at x=1. Does that mean that f also doesn't or to I have to check explicity with a limit? I think that I have to do a limit, is that right? Could someone elaborate on this?
    Thanks.
     
  2. jcsd
  3. Jul 30, 2007 #2

    Dick

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    The chain rule says (f(g(x))'=f'(g(x))*g'(x). f' may not have a limit but the product may. In your specific example it does. So yes, you have to check.
     
  4. Jul 30, 2007 #3

    daniel_i_l

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    Thanks. But you say that in this case f has a derivative at x=1? I did the limit and got infinity. Is that right?
     
  5. Aug 3, 2007 #4

    daniel_i_l

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    Anyone?
    Thanks.
     
  6. Aug 3, 2007 #5
    What is the derivative function you got? I got

    [tex]
    f'(x) = \frac{2-2x^2}{(1+x^2)\sqrt{1+2x^2-3x^4}}
    [/tex]

    If that is right, then the limit x->1 is non-trivial, but I think I got, using L'Hospital's rule, that it is zero and not infinity.
     
  7. Aug 3, 2007 #6

    George Jones

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    Factoring the top and bottom of your expression also shows that the "problem" disappears.
     
  8. Aug 3, 2007 #7

    Dick

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    I don't think that derivative is quite correct. After factoring what I get I'm just left with 2/(1+x^2).
     
  9. Aug 3, 2007 #8

    George Jones

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    Until now, I didn't actually do the derivative. I get (with a minus sign) what you have.
     
  10. Aug 3, 2007 #9
    I found the mistake, I had calculated [itex]1 + 2x^2 + x^4 - 4x^2 = 1 + 2x^2 - 3x^4[/itex]

    Now I got
    [tex]
    \frac{2}{1+x^2}
    [/tex]
    too, without minus sign though.
     
    Last edited: Aug 3, 2007
  11. Aug 4, 2007 #10

    George Jones

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    Actually, the sign is bit tricky, which I didn't see until I graphed

    [tex]arcsin\left(\frac{2x}{1+x^2}\right).[/tex]

    The derivative of this function has a jump discontinuity at [itex]x=1[/itex].

    Hint: consider

    [tex]\frac{-4}{\sqrt{\left(-4\right)^2}}.[/tex]
     
    Last edited: Aug 4, 2007
  12. Aug 4, 2007 #11
    I didn't check the domain earlier, and though that the function would be defined on some interval around origo, but actually the expression

    [tex]
    \frac{2x}{1+x^2}
    [/tex]

    doesn't get any values outside [itex][-1,1][/itex]. So the function becomes defined for all real numbers.

    I took a closer look at the derivative. Isn't it this?

    [tex]
    \frac{2(1-x^2)}{|1-x^2|(1+x^2)}
    [/tex]

    So there's plus sign when [itex]x\in ]-1,1[[/itex], minus sign when [itex]x\in ]-\infty,-1[\;\cup\; ]1,\infty[[/itex], and the derivative does not exist for [itex]x\pm 1[/itex].
     
  13. Aug 4, 2007 #12

    daniel_i_l

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    I took the limit of [tex]\frac{arcsin\left(\frac{2(x+1)}{1+(x+1)^2}\right) - \frac{\pi}{2}}{x}.[/tex] when x goes to 0 and got
    [tex]
    \frac{-2}{1+(x+1)^2}
    [/tex]
    By using L'hopital. It looks like what George got except that instead of x I have x+1. What did I do wrong?
    Thanks
     
  14. Aug 4, 2007 #13
    That limit should not exist. However if take only left handed or right handed limit then you should get either 1 or -1. Do you know which limit you are taking?

    If I got this right, then left handed limit should give 1, and right handed limit -1.

    I'm not sure if there is a mistake concerning the x+1 term. You must take the limit x->0 in the end, and then you also have x+1 -> 1. You have now inserted an expression x+1 in the place of the old parameter x, so x+1 -> 1 is precisely what you are supposed to have.
     
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