For my part, I think it would be easy if you do a procedure like this:

> test := proc (a0)

 local a, i, n;

 a[0] := a0;

 a[1] := 2+ln(a[0]);

 for i from 0 while 10^(-5) < abs(a[i]-a[i+1]) do

 a[i+2] := 2+ln(a[i+1]);

 n := i;

 end do;

 evalf(a[n], 10);

 end proc:

> test(1.0);

                                 3.146176412

Don't forget to give a0 in a decimal point.  And if you want to write the procedure, at the end of each line, do shift-enter.  You do enter only at the end to tell Maple to execute.  In this example, after end proc:

 I copy and paste the answers of dharr and everything works fine.  The only problem I could think is : an error of syntax or maybe you should do restart and than redo the calculation.

First of all, it is preferable to copy and paste the input and output of Maple directly in the post.  To do so, for example, if you have this in your worksheet, highlight the input and output then copy it and paste it in here.  Al you have to do after that is to erase the second line:

> eq1 := x^2+y(x)^2 = 1;
print(`output redirected...`); # input placeholder
                                     2    2   
                             eq1 := x  + y(x)  = 1
 

to respect the format, higlight again and chose the format from normal to formatted.  This is what you will have.

> eq1 := x^2+y(x)^2 = 1;

                                     2      2    
                             eq1 := x  + y(x)  = 1

This way, it is more easier to copy and paste for us to help you.

Concerning your problem,  Maple cannot plot if you define the dependant variable represent this way:, you have to do:

> eq2 := subs(y(x) = y, eq1);

                                     2    2    
                             eq2 := x  + y  = 1

Then you can use implicitdiff

> deq1 := implicitdiff(eq2, y, x);

                                           x
                                 deq1 := - -
                                           y

and now, you are ready to plot with implicitplot from the plots packages:

>with(plots)

> implicitplot(deq1=2, x = -2 .. 2, y = -2 .. 2);

Hope it help!

______________

 Thank you for your encouraging word.

I have been teaching Calculus for a while now.  But I have to follow a certain order.  Here, the students have 2 years wich give 4 semesters (of 4 month each)

1- differential calculus

2- integral calculus

3- probality and statistic

4- linear algebra (including complex numbers)

2, 3 and 4 is a piece of cake because they have the right background (1).  My goal was to find a way to use the formula for most of the time:

> Diff(f(x), x) = Limit((f(x+h)-f(x))/h, h = 0);

                      d                /f(x + h) - f(x)\
                     --- f(x) =  lim   |---------------|
                      dx        h -> 0 \       h       /

So here is what I do;
a) Concept of limit for simple case (no indetermination)
 and their properties
b) secante versus tangent
c) derivative of x^n
d) sum, product, quotient and chain rules
e) extremum of function, rates and differents application
f) derivative of ln(x), a^x, sin(x), implicit function and sandwich theorem etc
g) Hospital's rule and return on limit with indetermination
h) More applications (Newton, numeric solution of ODE etc)
i) parametric function and theirs deriva
tives

So my goal is to give rigourus but simple theoratical foundation
 of the derivative of functions.


After working on this, the only thing I have to work on is to prove that:
                               / h    \\
                               |a  - 1||
                         lim   |------||  =  ln(a)
                        h -> 0 \  h   //
  
without the use of table if possible.

About the comment you give me:

"Generally, after reading a few of your posts on this site, I would
recommend you to chose a specialty not related to mathematics in any
way."

Well, I don't teach at the university level but I think that
 I can do a good job at my level while using Maple, wich is not the cas for others
teachers.

So  unless you want to banning me from the site, I will continue to posts my question
in this site.

Mario

 All I want to do is construct the rule of differenting the functions all with the formula

> Diff(f(x), x) = Limit((f(x+h)-f(x))/h, h = 0);

                      d                /f(x + h) - f(x)\
                     --- f(x) =  lim   |---------------|
                      dx        h -> 0 \       h       /

So I find the rule for x^n. Then I establish the sum (minus) rule, the multipication rule, the quotient rule.

Then I show that the definition of "e" is

> Limit((1+h)^(1/h), h = 0) = e;

                                          /1\    
                                          |-|    
                                          \h/    
                             lim   (1 + h)    = e
                            h -> 0               

Then knowing that:

y = a^x and log[a](y) = x

find the derivative of both function with the definitions that I have establish and with the formula of the beginning.

As for sin(x), the student know that it's a periodic function F(x) = F(x + 2*Pi).  And then we construct the rule of derivative for cos(x), tan(x) and so on....

And finally the chain rule and l'Hospital's rule.  So now they have all they need to find the derivative of every kind of functions.

So you see that the derivative of a^x, ln(x) and sin(x) are the functions that I want to prove without using table of the fucntion while h tend to zero and I cannot use a series because it implie that I know the derivative of the fuction we are looking for.

I hope this will help you.

My goal, in teaching a first course of calculus, is to be able to calculate the derivative of the function a^x ,  ln(x) and sin(x)  with the definition:

> Diff(f(x), x) = Limit((f(x+h)-f(x))/h, h = 0);

                      d                /f(x + h) - f(x)\
                     --- f(x) =  lim   |---------------|
                      dx        h -> 0 \       h       /

First, I show that:

> Limit((1+h)^(1/h), h = 0) = e;

                                          /1\    
                                          |-|    
                                          \h/    
                             lim   (1 + h)    = e
                            h -> 0               

with a list that give the value of the limit as h approche 0.  I don't know any other way to demonstrate.

Then I look for a^x

> deq1 := a^x*(Limit((a^h-1)/h, h = 0));

                                   /       / h    \\
                                 x |       |a  - 1||
                        deq1 := a  | lim   |------||
                                   \h -> 0 \  h   //

I show that for a = 2, the limit is 0.6931 and that for a = 3, the limit is 1.0986.  So is there a function that the derivative is the function of herselft.  So we look at:

> Limit((a^h-1)/h, h = 0) = 1;

                                    / h    \    
                                    |a  - 1|    
                              lim   |------| = 1
                             h -> 0 \  h   /    
           

and the solution is a = e.  But for  a different value of a, I am stuck with:

> deq2 := a^x*(Limit((a^h-1)/h, h = 0));

                                   /       / h    \\
                                 x |       |a  - 1||
                        deq2 := a  | lim   |------||
                                   \h -> 0 \  h   //

Then I go with :

a = 2 in deq2 give 0.6931471806*2^x = ln(2)*2^x

a = 3 in deq2 give 1.098612289*3^x = ln(3)*3^x

And so on and I deduct that the result is in general:  I am not very please with that way of proving this.  So for  now, I have show that

> Diff(a^x, x); % = value(%);

                               d   x    x      
                              --- a  = a  ln(a)
                               dx              

Of course, I could use the following

> Diff(ln(a^x), x) = (Diff(a^x, x))/a^x;

                                           d   x
                                          --- a 
                              d    / x\    dx   
                             --- ln\a / = ------
                              dx             x  
                                            a   
> Diff(x*ln(a), x) = ln(a);

                             d                   
                            --- (x ln(a)) = ln(a)
                             dx                  

But to do that, I need to know the derivative of ln(x)

> Diff(Log[a](x), x) = Limit((log[a](x+h)-log[a](x))/h, h = 0);

                                        /ln(x + h)   ln(x)\
                                        |--------- - -----|
                  d                     |  ln(a)     ln(a)|
                 --- Log[a](x) =  lim   |-----------------|
                  dx             h -> 0 \        h        /


                                                   /  /x + h\\
                                                   |ln|-----||
                      /ln(x + h)   ln(x)\          |  \  x  /|
                      |--------- - -----|    lim   |---------|
                      |  ln(a)     ln(a)|   h -> 0 \    h    /
                lim   |-----------------| = ------------------
               h -> 0 \        h        /         ln(a)
and       

                                  /  /x + h\\    
                                  |ln|-----||    
                                  |  \  x  /|   1
                            lim   |---------| = -
                           h -> 0 \    h    /   x

wich I have to prove again.  I have a nice way for doing so:

 `Δy`/`Δx` = Log[a](1+`Δx`/x)/`Δx`;

                                        /    Delta;x\
                                  Log[a]|1 + --------|
                       Delta;y         \       x    /
                       -------- = --------------------
                       Delta;x         Δx      

Multiplying and dividing the last output and setting Delta;x/x

= alpha we have

> `Δy`/`Δx` = Log[a](1+alpha)^(1/alpha)/x;

                                                 /  1  \
                                                 |-----|
                                                 \alpha/
                     Δy   Log[a](1 + alpha)       
                     -------- = ------------------------
                     Δx              x            
and having prove (with a table of numbers) that

                                          /1\    
                                          |-|    
                                          \h/    
                             lim   (1 + h)    = e
                            h -> 0               

then

                           d              Log[a](e)
                          --- Log[a](x) = ---------
                           dx                 x    

And I prove the same way that

> Diff(sin(x), x) = cos(x)*(Limit(sin(h)/h, h = 0));

                     d                  /       /sin(h)\\
                    --- sin(x) = cos(x) | lim   |------||
                     dx                 \h -> 0 \  h   //

and that, with a Table;

                                    /sin(h)\    
                              lim   |------| = 1
                             h -> 0 \  h   /    

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.

Is doing so, only with table of numbers, am I prooving something?  Does anyone have a beautyfull way to go?

Thanks in advance

 you will run in the same problem:

> Diff(ln(x), x) = ln(Limit((1+alpha)^(1/alpha), alpha = 0))/x;

                              /                      /  1  \\
                              |                      |-----||
                              |                      \alpha/|
                            ln|   lim     (1 + alpha)       |
                 d            \alpha -> 0                   /
                --- ln(x) = ---------------------------------
                 dx                         x                
where alpha = (delta x)/x

You will have to prouve that the limit is "e" and that I can only show it
with a graph or a spreadsheet.


And in those last 3 examples, a graph or a spreadsheet is not a proof.

Is ther someone have an answer for me...?

in both case, you have to prove:

> deq1 := exp(x)*(Limit((exp(h)-1)/h, h = 0));

                                   /       /exp(h) - 1\\
                    deq1 := exp(x) | lim   |----------||
                                   \h -> 0 \    h     //

The liimit is 1.  But you have the result by the Hospital's Rule, wich you cannot do since it's what you are trying to prove.

> deq2 := a^x*(Limit((a^h-1)/h, h = 0));

                                   /       / h    \\
                                 x |       |a  - 1||
                        deq2 := a  | lim   |------||
                                   \h -> 0 \  h   //

While the limit is ln(a) by again using the Hospital's Rule.

For the moment, all I have to illustrate that the limit is ln(a) is to use the spreadsheet and show that the limit tend to be, for a = 5 for example, ln(5).  Of course, I would like a precises theoriticall.

 because we cannot see the exponent that you want to expand.

I realise that effectively, it was a circular demonstration and it was the reason why I did not used Hospital's Rule in the first place.  So the problem is still open!   Thanks Alex for pointing that out.

eq1 := y^3-a*y^2+y-a*b;

                                 3      2          
                         eq1 := y  - a y  + y - a b

Then, right-click to have the context menu and choose plot Builder -> Interactive Animation with 1 parameter and explore.

 How a calculator like TI83 calculate to give the answer of exp(5.2) or ln(3.4) etc.  For what I understand on this long exchange, is that it seems that there is a look-up table and with a hardware code, can give the exact answer starting with the table.  I was more interested in a simple way to show an exemple to the students.

I thought , and you show me that itI was false, that you could have a serie in the form of:

exp(n) = Sum(term containing "n") and that the more precise you want, you only have to take a further number of term.

But the level of your answer is far too high level for the students I was working with.

Anyway, thanks for those precises anwers.