Ok, it will ever be a matter of taste how to proceed, depending
on embedding and intention. And mine today would be not too much
puristic, something like done in Forster (how was/is with complex
Analysis) is fine for me.

Quickly introducing power series and exp is not so difficult and
the sketch of 2) or 3) above gives the derivative without much
pain.

Since monotony is evident you get log and after knowing you can
differentiate here its rule comes by 1 = (exp(log(x))' = x*log(x)'
by chain rule - or generally for differentiating an inverse.

However at some point one has to the work: he defines sin and cos
through exp(I*x) (yes, 1st semester), so the usual geometric view
has to be proven (no, not complex log, that is later and something
like in Cartan seems natural then).

More or less he quickly goes off the basic definition for f' and
one has the impression these lectures intend to enable physicist
quickly with their needs.


PS: the only book written by him that I do not like much is his
'Riemannian Surfaces'. It is quite 'dry' to read. The best I had
where notes on diff Geometry (Hodge stuff) in Italian, since he
wanted to examine that theme for PhD and it was easier to use
them than reading books.

Hm ... do you have now what you need or is something missing?
I lost the overview a bit ...

What are your definitions for the functions a^x ,  ln(x) and sin(x)?

"So I want to find the derivative of thoses 3 functions knowing only the derivative of x^n, the sum (minus) rule, the multipication rule, the quotient rule and the chain rule."

That sounds, as if you want to work on them as power series (+usual elementary diff properties).

What are your definitions for the functions a^x ,  ln(x) and sin(x)?

"So I want to find the derivative of thoses 3 functions knowing only the derivative of x^n, the sum (minus) rule, the multipication rule, the quotient rule and the chain rule."

That sounds, as if you want to work on them as power series (+usual elementary diff properties).

One could see the problems in the cited thread in a positive way:
Using the shown variants gives a special identity for some 2F1 :-)

Anyway, I give up (except you are willing (later) to show how you
came up with your task). Concerning x=1/6 the reformulation I got:

-25/343*42^(1/2)*x^(1/2)*
  hypergeom([-15/8, -1/2, -37/88],[-125/88, 1/8],-1/x)
+
81/539*42^(1/2)*
  hypergeom([-1/2, 3/8, 169/88],[81/88, 19/8],-x)
-
3/343*(6*x-1)*Pi^(3/2)*42^(1/2)*2^(1/2)/x^(11/8)/GAMMA(5/8)/GAMMA(7/8)
 
equals 0, so there is an analogous task for x=6.


I have no feeling, how difficult it is, but already Gamma's can be
a mess - here is one given by Bill Gosper: 

2^(1/4)*3^(3/8)*GAMMA(1/4)*GAMMA(1/3) = 
(3^(1/2)-1)^(1/2)*Pi^(1/2)*GAMMA(1/12)

yes, that's what I used in the above

but can not remember the prove of  ( exp(x) )' = exp(x)

think, it depends a bit on the defintions used, the easiest might be as analytic function, i.e. series

yes, that's what I used in the above

but can not remember the prove of  ( exp(x) )' = exp(x)

think, it depends a bit on the defintions used, the easiest might be as analytic function, i.e. series

has not worked for me, got just a double integral, it should stand for 3F2 =Int(2F1) = Int(Int(...)) using Euler's integral I guess ... and similar for trying sums :-(

and run into the 3rd oder DE (without splitting), it would not help to provide initial conditions here, but have not tried in points different from 0

BTW the task seems to stem from the problems seen in www.mapleprimes.com/forum/thomascalculus5615exercisegetshypergeom
where it was unclear how to simplify the 2F1

(a^x) ' = exp( ln(a)*x ) ' = exp( ln(a)*x ) * ( ln(a)*x ) '  =

= exp( ln(a)*x ) * ln(a) = a^x * ln(a) ---> ln(a) 

no?

(a^x) ' = exp( ln(a)*x ) ' = exp( ln(a)*x ) * ( ln(a)*x ) '  =

= exp( ln(a)*x ) * ln(a) = a^x * ln(a) ---> ln(a) 

no?

Using an arbitrary upper bound and specializing later works:

  v:=int(sqrt(t^5+6*t)*(5*t^4+6), t=0..b):
  eval(v,b=1);
                                   1/2
                               14 7
                               -------
                                  3

Using an arbitrary upper bound and specializing later works:

  v:=int(sqrt(t^5+6*t)*(5*t^4+6), t=0..b):
  eval(v,b=1);
                                   1/2
                               14 7
                               -------
                                  3

Dave, so what is your opinion - and those of your colleagues?

dito on Win XP Home SP2

plot(1/(cos(x)^2), x=0 .. 5*Pi*(1/4), y= 0..6);