Math Is Fun Forum

Ok, so I'm onto my second attempt... the first idea of ignoring the absolute values around the sin and cos when differentiating was WRONG.  So, I did it again, using the following:
1) the top-most item is to use the quotient rule, with cos(a) above the line and (|cos a|^4+...)^0.25 below the line
2) the chain rule is applied with fourth-root being the f and (|cos a|^4+|sin a|^4) being the g in
(fog)' = (f'og)g'
3) the sum rule takes care of |cos a|^4 + |sin a|^4
4) the chain rule takes care of |cos a|^4 where f is to-the-fourth and g is |cos a|, again with (fog)' = (f'og)g'
5) the chain rule takes care of |cos a| where f is absolute and g is cos(a)
4bis) and 5bis) do for |sin a|^4 in the same manner as |cos a|^4

I get a real-headache result, which I may try to LaTeX for you all.

I was pleased when I used Excel to differentiate the c(a) taking Δx as 0.01 radians, then plotting the graph.  The graph of d/dx(c(a)) from excel was excitingly the same shape as I got when I put my headache into graphcalc.

Back to my original post: "Does anyone know of the way to differentiate and integrate the functions?"  I can begin to differentiate the function given by Kurre using product, quotient and chain rules, but I will have to nest them and I will need a BIG piece of paper to deal with all the 'recursion' or 'nesting' of pieces in each other.

I will start with quotient rule: [u/v]' = (u'v - uv') / v^2
where u = cos(a) and v = (cos^n(a) + sin^n(a))^(1/4)
(I'm aware of ditching the unsigned or absolute, but I'm doing that for simplicity, and will limit my results to the a from 0 to pi/2)

If anyone gets there with a differential then I'd be interested to see it.  I suspect that a few trig simplifications which I don't know may result in a much tidier formula.

Go on, you know you want to...

By the use of chain rule dy/dx = dy/dh . dh/dx and also by writing
y = sqrt (1 - x^2)
I got
dy/dx = (-x) / sqrt(1-x^2)
which satisfied the trig rules, ie I had found an equivalent of -cot(alpha).

By the chain rule on x^4+y^4=1  I had
y=sqrt(sqrt(1 - x^4))
which differentiated to
dy/dx = (-4 x^3) / sqrt(sqrt((1 - x^4) ^3))
and seems to give a gradient on the tanget on my lines when I plot on Excel.

Looking at wikipedia (see http://en.wikipedia.org/wiki/Trig_functions) I see what remember,
sin x = x  - x^3/3! + x^5/5! - ...
and I have wondered if there can be a similar infinite series for the X^4 form I've considered.

I looked at the Co-tangent observations made by Kurre (above) but couldn't follow his (or, her) step from
(y*CotA)^n+y^n=1
to
s^n(A)=y=1/(Cot^nA+1)
So thanks for the reply but sorry I didn't "get" it.