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Dy/dx of arcsin(1/x^4) clarifying question

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Find dy/dx of arcsin(1/x^4)



    2. The solution

    Answer simplified:

    -4/(x*sqrt(x^8-1))

    I've checked my answer to the above problem on WolframAlpha, and Wolfram states that the answer is "an alternate form assuming x is positive"

    I guess this is more of a clarifying question, but what does Wolfram mean by that? As far as I understand, x doesn't need to be positive in order for the simplified solution to remain valid.
     
  2. jcsd
  3. Nov 8, 2012 #2

    Simon Bridge

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    What would the solution be if it wasn't simplified?
    What form did Wolfram consider this to be an alternate to?
     
  4. Nov 10, 2012 #3
    Hello Simon,

    Thanks for your reply. The form Wolfram gave me was:

    -4/[sqrt(1-(1/8))*(x^5)]

    http://www.wolframalpha.com/input/?i=dy/dx+of+arcsin(1/x^4)

    I still don't see where the 'simplified form' that I derived and this form differ.
     
  5. Nov 10, 2012 #4

    Simon Bridge

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    "simplified" term -4/(x*sqrt(x^8-1)):$$y_s=-\frac{4}{x\sqrt{x^8-1}}$$ unsimplified term -4/[sqrt(1-(1/8))*(x^5)]
    should be: $$y=-\frac{4}{x^5 \sqrt{1-\frac{1}{x^8}}}$$... perhaps: ##\sqrt{x^8}=\pm x^4##
     
  6. Nov 10, 2012 #5
    Could you please elaborate on this?

    Please correct me if I'm wrong, but I think that statement would be correct only if it were ##\pm\sqrt{x^8}##, but it's only ##\ +\sqrt{x^8}##
    Plus, how did you isolate ##\sqrt{x^8}## from ##\sqrt{x^8-1}##?
     
  7. Nov 10, 2012 #6
    Perhaps Alternate form assuming x is positive doesn't mean that the input x has to be positive. Maybe it's referring to the function as a whole (or something else)?

    Wolfram says the samething when I input:

    ##\sqrt{x^8}##

    and that ##x^8## is an alternate form assuming x is positive, but we know that ##x^8## behaves exactly the same as ##\sqrt{x^8}## if the input x is negative.
     
  8. Nov 10, 2012 #7
    Last edited: Nov 10, 2012
  9. Nov 10, 2012 #8

    SammyS

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    That's strange.

    WolframAlpha says that [itex]\displaystyle \ \ -\frac{4}{x^5 \sqrt{1-\frac{1}{x^8}}}+\frac{4}{x\sqrt{x^8-1}}=0\ \ [/itex] if x > 0 .

    Added in Edit:

    You have typos in those expressions in the links to WolframAlpha.
     
    Last edited: Nov 10, 2012
  10. Nov 10, 2012 #9

    Simon Bridge

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    Wolfram is accounting for the possibility that ##x^4<0##

    You started with:$$y(x)=\arcsin(x^{-4}) \Rightarrow -1 \leq x^{-4} \leq 1$$...for this relation to be meaningful, y' will only be real for x=±1.
    Hence, you get different numbers depending on whether ##x^4## is positive or negative. Try it for some value of x?
    [edit: darn - seems to work out the same if I just make x<0 - but I suspect something like this is what is going on]
     
    Last edited: Nov 10, 2012
  11. Nov 10, 2012 #10

    SammyS

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    x4 is non negative for real x.

    [itex]\sqrt{x^8}[/itex] is non negative for real x.

    [itex]\sqrt{x^8}=x^4[/itex] for all real x.
     
  12. Nov 10, 2012 #11

    Simon Bridge

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    But how is wolfram supposed to know that the input is restricted to real numbers only?

    ... looking over the several examples in wolfram again - that is exactly what they are doing.
    I think the entry is mislabelled in these instances: in every case, the alternate is the form assuming x is REAL.

    technically, Wolfram is evaluating ##\text{sqrt(x)}## not ##\sqrt{x}## ... so

    $$\sqrt{x^2}=xe^{i\pi \lfloor \frac{1}{2}-\frac{\arg(x)}{\pi} \rfloor}$$ this becomes ##x## in the case that ##x^2\geq 0##
     
    Last edited: Nov 10, 2012
  13. Nov 14, 2012 #12
    I'm sorry, but I haven't learned this yet, but I think you guys are saying that Wolfram should have labeled it an "alternate form assuming x is real" (rather than just positive).

    Excuse my ignorance, but what's the difference and how does this relate to Wolfram's statement that x has to be positive for the alternate form?

    Thanks!
     
  14. Nov 14, 2012 #13

    Simon Bridge

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    The map is not the territory.

    ##\text{sqrt(x)}## is an algorithm defined in a computer program, in this case: mathematica, while ##\sqrt{x}## is a mathematical statement defined according to the rules of arithmetic. Computer programs play by whatever rules the programmer gives them.

    One of the rules is what to say when a simplification is not always valid.

    That is what I am saying ... however, it may be tricky to program it that way and also allow for other possibilities. The situation becomes clear upon examination after all.
    This sort of thing can happen when you become aware of more possibilities than the person asking the question. Perhaps you'd like to submit it as a bug?
     
    Last edited: Nov 14, 2012
  15. Nov 15, 2012 #14

    Mark44

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    A nit: You don't "find dy/dx" of something; dy/dx already is the derivative of y with respect to x.

    When you learn about implicit differentiation it will be important to distinguish between a pending operation of differentiation (such as d/dx(x3 + xy2) and something that already is a derivative (such as dy/dx or y').
     
  16. Nov 15, 2012 #15

    Simon Bridge

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    Nit picking can be fun... however, you should have provided the correct, non-nit-pick-attracting, form:

    Technically, it should be: "find dy/dx where y=arcsin(1/x^4)" right?
    Still, though "dydx" is logically the correct answer - it would not get many marks. To completey avoid nit-picking you'd have to write it out longhand in full.

    eg. "find the derivative, with respect to x, of y=x^2"
    ...would return "d(y=x^2)/dx <=> y'=2x"
    ... though I bet it is possible to nit-pick that one too :)

    ... but since this is the English language and not a logic thesis, the "of", in OPs remark, can easily stand for "where y=" with the intention provided by context.

    You did say it was a "nit pick" - so it is a nit pick for me to point out that most people familiar with contextual languages also know to modify their language when an ambiguity is possible. OP will figure this out when implicit differentiation is learned - if it hasn't already.

    Reminds me of the exam question "find x" it says ... the student carefully circles the "x" and writes "there it is".
     
  17. Nov 15, 2012 #16
    Thanks for all your help Simon (and others)! I think I'll rest with the fact that it's something to do with WolfRam's programming.
     
  18. Nov 16, 2012 #17

    Mark44

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    The thought actually did cross my mind, but I decided that it was enough to point out the mistake of "find dy/dx of ..." without having to also restate the problem.
    Or more briefly, "find dy/dx if y = x2."
    Particularly the part about taking the derivative of an equation rather than of a function.
    I suppose that could happen, but in my experience of teaching calculus for more than 20 years, many students are unclear about the distinction between the operator d/dx and the function dy/dx. This was what I was trying to point out to the OP.
    This is really a stretch, IMO. I believe that one goal in mathematics teaching is to promote a style of writing, thinking, and speaking with a minimum of ambiguity. Students already have a difficult time trying to comprehend abstract concepts, even when they are presented in a straightforward manner with no ambiguity.
     
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