```Hi Johann and Josef,

of course I was wrong in writing f'(x) instead of f'(a), but only in the
mail .

And of course one should have an open 2D-plot to get cursors coordinates.
But fact is, that in Vers.5.04 you don´t have to simplify first but in
Vers.5.06
I sprung over V5.05, so I can't tell when this behaviour changed.
AND WHY? (Ask Theresa and Al)
Johann made the problem clear: by substitute a=hCross BEFORE simplifying you
get
the result I copied in line #9.
Ok, Johann is right again when he mentioned that simplifcation BEFORE using
should
be done anyway.
BUT in didactic matters in the beginning I won´t have my students have a
look on
the simplified form. Later on when they are used to work with DERIVATES its
ok.

Let me mention another visualisation aspect:
you can have the function-plot invisible in the background color and even
the cursor
invisible while in trace mode you plot the tangents
and ask the students for the type of the function.
But don´t forget to take another plotcolor for the lines.

Rainer

> -----Ursprüngliche Nachricht-----
> Von: DERIVE computer algebra system
> Gesendet: Samstag, 25. Januar 2003 16:09
> Betreff: Re: AW: Derivates in Derive 5.06
>
>
> Hi Josef,
>
> Let me just add some remarks of my own on Rainer's problem.
> In the first
> place, as you already mentioned, he made a mistake in the
> second line and
> here is the correct version of his program. (Note that Derive
> won't accept
> the third line unless you open the 2D-plot window and set the
> mouse cursor
> somewhere!)
>
>
> f(x) := x^2 - 5x + 7
> tang(x, a) := f'(a)(x - a) + f(a)
> tang(x, hCross)
>
> This will not work though, as there will be the notorious
> message "Sorry,
> the highlighted expression cannot be plotted!" But there is a
> simple and
> straight-forward remedy, which is: "Simplify the second
> line!". Obviously,
> Rainer's pupils did this at school and he forgot (?) to do
> this at home.
>
> In the same way another more serious problem can be fixed.
> Let's define the
> following function
>
> f1(x):=dif(fx,x)
>
> I know this is not necassary as all derivatives f'(x),f''(x), etc. are
> available in this form anyway. (Believe it or not, before
> mail I had not been aware of this fact!) Anyway,...If we try
> to find the
> value of f1(x) at x=2 we get
>
> f1(2) = DIF(1, 2)
>
> It should be clear what went wrong here: Derive substitutes
> x=2 before (!)
> forming the first derivative. (Sadly enough, this deficiency
> was introduced
> in the most recent version 5.06!) Again, simplifying the definition of
> f1(x) before using it will help. (This should be done in this
> case anyway
> and that's why only novices will ever encounter this error!)
> On the other
> hand, the simplification of f'(2) doesn't cause any problems
> at all, which
> is really strange!
>
> Hope these remarks helped to throw more light upon the matter.
>
> Cheers,
> Johann
>
>
>
> It deserves to be mentioned that
> At 10:09 AM 1/25/03 +0100, you wrote:
> >Content-Type: text/plain;
> >        charset="iso-8859-1"
> >X-MIME-Autoconverted: from 8bit to quoted-printable by
> ori.rl.ac.uk id
> h0P92cS21938
> >
> >Dear all (an special greetings to my friend Heinz-Rainer).
> >
> >As I can see he could save his PC of the daring floods in his home.
> >I appreciate your application of this "hidden" feature of
> DERIVE and I am
> >sure that there are many other ways for a meaningful use of
> (hcross,vcross).
> >
> >
> >(1) I believe there is a typing error in the second line of
> >
> >tang(x,a):=f'(a)*(x-a)+f(a). But this was not the problem.
> >
> >You can overcome the difficulty by first deriving and then
> evaluating the
> >derivative. (It is not as simple as on the TI-92 - and I
> >prefer the more mathematical way as it is performed be DERIVE.
> >
> >f(x) := x^2 - 5·x + 7
> >
> >tang(x, a) := LIM(DIF(f(x), x), x, a)*(x - a) + f(a)
> >
> >tang(x, hCross)
> >
> >This works properly. You might prefer a shorter version:
> >
> >tg := LIM(DIF(f(x), x), x, hCross)·(x - hCross) + f(hCross)
> >
> >each call of tg in the 2D-Plot Window returns a tangent of
> function f(x)
> >defined above. You only have to move the cursor cross in any
> position and
> >the horizontal coordinate of the cross places the respective tangent.
> >
> >tgpt := [tg, [hCross, f(hCross)]]
> >
> >returns not only the tangent but also the osculating point
> (set points
> >LARGE). You could go on and add the slope-triangle, the
> respective point to
> >find point by point the first derivative .....
> >
> >I attach the dfw-file with the graphs.
> >
> >Once more many thanks for pointing at this interesting
> possibility which
> >helps again visualizing mathematics.
> >
> >Best regards
> >Josef
> >
> >
> >-----Ursprüngliche Nachricht-----
> >Von: DERIVE computer algebra system
> >Gesendet: Freitag, 24. Januar 2003 21:39
> >Betreff: Derivates in Derive 5.06
> >
> >
> >Hi all, hi Josef (Böhm),
> >
> >i like the sketchpad idea in last DNL 48,p 32.
> >I tried to show moving tangents on some curves.
> >It all worked fine in school with Version 5.04
> >for example:
> >f(x):= x^2-5x+7
> >tang(x,a):= f'(x)*(x-a)+f(a)
> >tang(x, hCross)
> >
> >But when i repeated it at home with Version 5.06 i got the message:
> >Derive can´t plot that expression.
> >
> >I simplified tang(x,hCross) and receive:
> >                 |     57 |                 | 349    57 |    349
> >#9:              ¦x - ————¦·    lim      DIF¦—————, ————¦ + —————
> >                 |     20 | 57/20->57/20    | 400    20 |    400
> >
> >What has changed between 5.04 to 5.06?
> >
> >To get a result you have to simplify the function tang(x,a) first and
> >make a new definition
> >
> >                                    2
> >#14:  tang2(x0) := x·(2·x0 - 5) - x0  + 7
> >
> >Someone has an idea?
> >
> >Rainer
> >
> >Attachment Converted: "c:\internet\eudora\attach\heinz_rainer.dfw"
> >
>
```