Complex Hyperbolic Derivative Problem

[Liquid7800]

Homework Statement

Hello,
Thanks for taking some time to help me out...and I have to apologize for posting a graphic of my logic and attempted answer instead of using LateX (It would take me a very long time just to get this problem viewable)

Please help me check my work and logic here...(see attached graphic) I don't know of any place I can see if my answer is correct (Not a book problem)

The problem itself is stated in the first box drawn (see below)...

Homework Equations

None

The Attempt at a Solution

My logic (with comments throughout) and the last red box drawn is my attempt at a solution..(see the graphic below)
I hope this is sort of easy to follow ...and I appreciate any help.
Thank you...

Does this seem right?

 

[HallsofIvy]

No, it doesn't seem right. At one point you have
$$\frac{1}{2}\frac{1}{2}sech^2(\frac{x}{2})- \frac{1}{2}tanh(\frac{x}{2})sech(\frac{x}{2})$$ and then, in the next two lines, that difference has metamorphised into a product,
$$\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)\left(sech^2(\frac{x}{2})\right)\left(-1\right)\left(tanh^2(\frac{x}{2})\right)\left(sech^2(\frac{x}{2})\right)$$
and then in the next line, you have reassembled that difference into a sum!

The first line I mention above,
$$\frac{1}{2}\frac{1}{2}sech^2(\frac{x}{2})- \frac{1}{2}tanh(\frac{x}{2})sech(\frac{x}{2})$$
is correct. You can factor (1/2)sech²(x/2) out:
$$\frac{1}{2}sech^2(x/2)\left(\frac{1}{2}- tanh^2(\frac{1}{2})\right)$$
but that's about all you can do.

 

[Liquid7800]

@HallOfIvy..

Thanks very much for reviewing my work!

--------------------------------------------------------------------------------
No, it doesn't seem right. At one point you have
$$\frac{1}{2}\frac{1}{2}sech^2(\frac{x}{2})- \frac{1}{2}tanh(\frac{x}{2})sech(\frac{x}{2})$$
and then, in the next two lines, that difference has metamorphised into a product,
$$\frac{1}{2}\frac{1}{2}sech^2(\frac{x}{2})- \frac{1}{2}tanh(\frac{x}{2})sech(\frac{x}{2})$$
and then in the next line, you have reassembled that difference into a sum!

I was very suspicious of that operation (I thought it might be illegal)...thats why I was hoping an expert such as yourself could give me some insight.

Now I've got two questions:

1) The sbove mentioned operstion is illegal why?
What I was trying to do was turn the difference into a sum by factoring out the negative -1?
In effect turning function-function*function into (function+function)-1*function...
Or is this totally illegal...

2) So the final answer is:

$$\frac{1}{2}sech^2(x/2)\left(\frac{1}{2}- tanh^2(\frac{1}{2})\right)$$

or should it be:
$$\frac{1}{2}sech^2(x/2)\left(\frac{1}{2}- tanh^2(\frac{x}{2})\right)$$
...and so one of these is then the FINAL answer?