Results 1 to 3 of 3

Thread: Differentiating a function of hyperbolic functions

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    36

    Differentiating a function of hyperbolic functions

    I don't know how to show this:

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

    Then
    $\displaystyle x'_{1}(s)$$\displaystyle = \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.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003
    Quote Originally Posted by bobby View Post
    I don't know how to show this:

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

    Then
    $\displaystyle x'_{1}(s)$$\displaystyle = \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 "$\displaystyle 2\frac{s}{4}$"? That is, of course, just $\displaystyle \frac{s}{2}$ but I suspect that the "2" is not supposed to be there.

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

    Now, $\displaystyle \frac{1}{2}\frac{1}{\sqrt{1- \frac{s^2}{16}}}- \frac{s^2}{8}\frac{1}{1-\frac{s^2}{16}}$$\displaystyle = \frac{1}{2}\frac{1- \frac{s^2}{16}}{\sqrt{1- \frac{x^2}{16}}}$$\displaystyle = \frac{1}{2}\sqrt{1- \frac{s^2}{16}}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    36
    Quote Originally Posted by HallsofIvy View Post
    Is that really "$\displaystyle 2\frac{s}{4}$"? That is, of course, just $\displaystyle \frac{s}{2}$ but I suspect that the "2" is not supposed to be there.

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

    Now, $\displaystyle \frac{1}{2}\frac{1}{\sqrt{1- \frac{s^2}{16}}}- \frac{s^2}{8}\frac{1}{1-\frac{s^2}{16}}$$\displaystyle = \frac{1}{2}\frac{1- \frac{s^2}{16}}{\sqrt{1- \frac{x^2}{16}}}$$\displaystyle = \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 $\displaystyle \frac{s}{4}$ 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 $\displaystyle \frac{s^{2}}{8}$ in the square root, should this not be over 16?

    Secondly,

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

    $\displaystyle \frac{1}{2}\frac{1}{\sqrt{1- \frac{s^2}{16}}}$ is obvious but the other part is not so clear.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differentiating some functions.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 17th 2011, 04:07 AM
  2. Differentiating sin functions
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Mar 27th 2011, 04:16 PM
  3. differentiating log functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 19th 2009, 04:15 PM
  4. differentiating a hyperbolic function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 18th 2009, 01:22 PM
  5. differentiating 3 functions
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Apr 15th 2009, 03:12 PM

Search Tags


/mathhelpforum @mathhelpforum