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