# Differentiating a function of hyperbolic functions

• May 20th 2009, 12:49 PM
bobby
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.

• May 20th 2009, 01:32 PM
HallsofIvy
Quote:

Originally Posted by bobby
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, $\frac{1}{2}\frac{1}{\sqrt{1- \frac{s^2}{16}}}- \frac{s^2}{8}\frac{1}{1-\frac{s^2}{16}}$ $= \frac{1}{2}\frac{1- \frac{s^2}{16}}{\sqrt{1- \frac{x^2}{16}}}$ $= \frac{1}{2}\sqrt{1- \frac{s^2}{16}}$
• May 20th 2009, 02:37 PM
bobby
Quote:

Originally Posted by HallsofIvy
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.