How to plot derivative directly?
Šerých Jakub
Dear mathgroup,
it seems to me, that response to my question shall be very simple,
but I cannot find it. :-(
I want to plot the derivative of the function. I would like to do it
directly, something like:
Plot[D[x^3 - 6 (x + 1)^2 + x - 7, x],{x,-3,8}]
It returns: General::ivar: "-2.99978 is not a valid variable."
I can understand that it is because local variable x from Plot command
interferes with the x variable from the D[].
Yes I can bypass the problem by:
deriv = D[x^3 - 6 (x + 1)^2 + x - 7, x]
Plot[deriv, {x, -3, 8}]
which is fully functional, but as far as I know Mathematica, there must
be some simple solution how to do it directly inside the Plot[].
Thanks in advance for kick-off
Jakub
Murray Eisenberg
of x to its first argument, you are now trying to take the derivative of
a constant with respect to a constant.
Another way to do it is:
f[x_]:= x^3 - 6 (x + 1)^2 + x - 7
Plot[f'[x],{x,-3,8}]
Or, perhaps a bit better f the function is much more complicated:
Plot[Evaluate[f'[x],{x,-3,8}]
--
Murray Eisenberg mur...@math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
David Skulsky
David
Peter Pein
Hi Jakub,
have a look at
http://reference.wolfram.com/mathematica/ref/HoldFirst.html (an
attribute being set for Plot) and use
Plot[D[f[x],x]//Evaluate,{x,x0,x1}]
Peter
Helen Read
Well, you can do this:
Plot[Evaluate[D[x^3 - 6 (x + 1) 2 + x - 7, x]], {x, -3, 8}]
But I think it's worth defining your function as a function.
g[x_] := x^3 - 6 (x + 1) 2 + x - 7
Plot[g'[x], {x, -3, 8}]
--
Helen Read
University of Vermont
DrMajorBob
or
Clear[f]
f[x_] = x^3 - 6 (x + 1)^2 + x - 7;
Plot[f'[x], {x, -3, 8}]
Bobby
On Mon, 11 Apr 2011 06:05:40 -0500, =A6er=FDch Jakub <Ser...@panska.cz> wrote:
>
> Dear mathgroup,
>
> it seems to me, that response to my question shall be very simple,
> but I cannot find it. :-(
>
> I want to plot the derivative of the function. I would like to do it
> directly, something like:
>
> Plot[D[x^3 - 6 (x + 1)^2 + x - 7, x],{x,-3,8}]
>
> It returns: General::ivar: "-2.99978 is not a valid variable."
>
> I can understand that it is because local variable x from Plot command
> interferes with the x variable from the D[].
>
> Yes I can bypass the problem by:
> deriv = D[x^3 - 6 (x + 1)^2 + x - 7, x]
> Plot[deriv, {x, -3, 8}]
>
> which is fully functional, but as far as I know Mathematica, there must
> be some simple solution how to do it directly inside the Plot[].
>
> Thanks in advance for kick-off
>
> Jakub
>
Bill Rowe
>it seems to me, that response to my question shall be very simple,
>but I cannot find it. :-(
>I want to plot the derivative of the function. I would like to do it
>directly, something like:
>Plot[D[x^3 - 6 (x + 1)^2 + x - 7, x],{x,-3,8}]
>It returns: General::ivar: "-2.99978 is not a valid variable."
>I can understand that it is because local variable x from Plot
>command interferes with the x variable from the D[].
More specifically, Plot substitutes a numerical value for x
every where it appears in the expression then evaluates the
expression. The way to plot the derivative is to force
evaluation of the derivative before Plot substitutes a numerical
value for x. That is you want to do:
Plot[Evaluate[D[x^3 - 6 (x + 1)^2 + x - 7, x]], {x, -3, 8}]
And even if you could somehow get
Plot[D[x^3 - 6 (x + 1)^2 + x - 7, x],{x,-3,8}]
to work, you really wouldn't want to do that since it would mean
re-evaluating the derivative for each numerical value of x. That
would require a lot of unneeded computation.
Stefan
Jakub, the Plot[] function picks a set of discrete x values (within
the range specified, very closely spaced to create the smooth graph)
and tries to evaluate your function with those numbers in place of the
x's. This creates a problem when it then tries to evaluate
D[-3^3-6(-3+1)^2+(-3)-7, -3], since you cant differentiate with
respect to a number like -3, hence the 'General::ivar: "-2.99978 is
not a valid variable." ' error message. You can get it to evaluate the
derivative before substituting in numbers for x by simply enclosing
your derivative inside an Evaluate[]
Plot[Evaluate[D[x^3 - 6 (x + 1)^2 + x - 7, x]], {x, -3, 8}]
Hope that helps!
-Stefan
Anthony Hodsdon
--Anthony
Wolfgang Windsteiger
this is due to
In[3]:= Attributes[Plot]
Out[3]= {HoldAll, Protected}
You must force Mathematica to evaluate the first parameter of Plot
before Plot goes on processing. The easiest way to accomplish this is an
explicit "Evaluate":
Plot[Evaluate[D[x^3 - 6 (x + 1) 2 + x - 7, x]], {x, -3, 8}]
Hope this helps,
ciao,
WW.
On 04/11/2011 01:05 PM, Šerých Jakub wrote:
> Plot[D[x3 - 6 (x + 1)2 + x - 7, x],{x,-3,8}]
Alexei Boulbitch
might this:
Plot[D[x^3 - 6 (x + 1)^2 + x - 7, x] // Evaluate, {x, -3, 8}]
be what you are looking for?
Have fun, Alexei
Dear mathgroup,
it seems to me, that response to my question shall be very simple,
but I cannot find it. :-(
I want to plot the derivative of the function. I would like to do it
directly, something like:
Plot[D[x^3 - 6 (x + 1)^2 + x - 7, x],{x,-3,8}]
It returns: General::ivar: "-2.99978 is not a valid variable."
I can understand that it is because local variable x from Plot command
interferes with the x variable from the D[].
Yes I can bypass the problem by:
deriv = D[x^3 - 6 (x + 1)^2 + x - 7, x]
Plot[deriv, {x, -3, 8}]
which is fully functional, but as far as I know Mathematica, there must
be some simple solution how to do it directly inside the Plot[].
Thanks in advance for kick-off
Jakub
--
Alexei Boulbitch, Dr. habil.
Senior Scientist
Material Development
IEE S.A.
ZAE Weiergewan
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L-5326 CONTERN
Luxembourg
Tel: +352 2454 2566
Fax: +352 2454 3566
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e-mail: alexei.b...@iee.lu
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Bob Hanlon
f[x_] = x^3 - 6 (x + 1)^2 + x - 7;
Plot[f'[x], {x, -3, 8}]
Attributes[Plot]
{HoldAll, Protected}
Plot[Evaluate[D[f[x], x]], {x, -3, 8}]
Bob Hanlon
---- "=C5 er=C3=BDch Jakub" <Ser...@panska.cz> wrote:
==========================
Peter Breitfeld
> Plot[D[x^3 - 6 (x + 1)^2 + x - 7, x],{x,-3,8}]
Use Evaluate:
Plot[D[x^3 - 6 (x + 1)^2 + x - 7, x] // Evaluate, {x, -3, 8}]
--
_________________________________________________________________
Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de
DC
Richard Hofler
In[9]:= deriv=D[x^3-6 (x+1)^2+x-7,x];
plot1 = Plot[deriv,{x,-3,8}]
An alternative method with Evaluate inserted:
In[11]:= plot2 = Plot[Evaluate[D[x^3-6 (x+1)^2+x-7,x]],{x,-3,8}]
The two plots are the same.
In[12]:= plot1==plot2
Out[12]= True
Richard Hofler
________________________________________
From: =8Aer=FDch Jakub [Ser...@panska.cz]
Sent: Monday, April 11, 2011 7:05 AM
Subject: How to plot derivative directly?
Dear mathgroup,
Heike Gramberg
with Attributes[Plot]). This means that the derivative doesn't get evaluated until
after w has been replaced with numerical values. To overcome this, you need to
tell mathematica to evaluate the derivative before plugging in numbers. This can
be done with evaluate:
Plot[Evaluate[D[x^3 - 6 (x + 1)^2 + x - 7, x]], {x, -3, 8}]
Heike
papu...@gmail.com
Just use a very simple command