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Math Help - Differentiating a function of hyperbolic functions

  1. #1
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    Differentiating a function of hyperbolic functions

    I don't know how to show this:

    If  x_{1}(s) = 2\arcsin(\frac{s}{4}) + 2\frac{s}{4}\sqrt{1-\frac{s^{2}}{16}}

    Then
     x'_{1}(s)  = \frac{1}{4}\sqrt{16-s^{2}}

    Can anyone help? Thanks.

    Obviously you are just differnetiating first using the rule for differentiating arcsin but then once I evaluate it all I do not get the required solution.

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  2. #2
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    Quote Originally Posted by bobby View Post
    I don't know how to show this:

    If  x_{1}(s) = 2\arcsin(\frac{s}{4}) + 2\frac{s}{4}\sqrt{1-\frac{s^{2}}{16}}

    Then
     x'_{1}(s)  = \frac{1}{4}\sqrt{16-s^{2}}

    Can anyone help? Thanks.

    Obviously you are just differnetiating first using the rule for differentiating arcsin but then once I evaluate it all I do not get the required solution.

    Is that really " 2\frac{s}{4}"? That is, of course, just \frac{s}{2} but I suspect that the "2" is not supposed to be there.

    If it were just  x_{1}(s) = 2\arcsin(\frac{s}{4}) + \frac{s}{4}\sqrt{1-\frac{s^{2}}{16}} then x_{1}'(x)= 2\frac{1/4}{\sqrt{1- \frac{s^2}{16}}}+ \frac{1}{4}\sqrt{1- \frac{s^2}{16}}+ \frac{s}{8}\frac{1}{\sqrt{1- s^2/8}}(-\frac{s}{8})

    Now, \frac{1}{2}\frac{1}{\sqrt{1- \frac{s^2}{16}}}- \frac{s^2}{8}\frac{1}{1-\frac{s^2}{16}} = \frac{1}{2}\frac{1- \frac{s^2}{16}}{\sqrt{1- \frac{x^2}{16}}} = \frac{1}{2}\sqrt{1- \frac{s^2}{16}}
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Is that really " 2\frac{s}{4}"? That is, of course, just \frac{s}{2} but I suspect that the "2" is not supposed to be there.

    If it were just  x_{1}(s) = 2\arcsin(\frac{s}{4}) + \frac{s}{4}\sqrt{1-\frac{s^{2}}{16}} then x_{1}'(x)= 2\frac{1/4}{\sqrt{1- \frac{s^2}{16}}}+ \frac{1}{4}\sqrt{1- \frac{s^2}{16}}+ \frac{s}{8}\frac{1}{\sqrt{1- s^2/8}}(-\frac{s}{8})

    Now, \frac{1}{2}\frac{1}{\sqrt{1- \frac{s^2}{16}}}- \frac{s^2}{8}\frac{1}{1-\frac{s^2}{16}} = \frac{1}{2}\frac{1- \frac{s^2}{16}}{\sqrt{1- \frac{x^2}{16}}} = \frac{1}{2}\sqrt{1- \frac{s^2}{16}}
    I'm reading it from a past paper and it does have the '2' in front of the \frac{s}{4} strangely. Either way, i'm not sure on a couple of parts of your method.

    Firstly,

    In the 1st line of the differentiation, the 3rd term, I am unsure of why there is an \frac{s^{2}}{8} in the square root, should this not be over 16?

    Secondly,

    I do not understand how you went from the first line to \frac{1}{2}\frac{1}{\sqrt{1- \frac{s^2}{16}}}- \frac{s^2}{8}\frac{1}{1-\frac{s^2}{16}}

    \frac{1}{2}\frac{1}{\sqrt{1- \frac{s^2}{16}}} is obvious but the other part is not so clear.
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