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 Index > Heap > Total Derivative and 3d Grapher
Author
Tyler

Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
1) What's a good 3d graphing program? Preferably GUI and open source.

2) I'll have to show you:
Code:
```f(x, y) = 5x^2 - 3xy + y^2
d/dx f = 10x - 3y + dy/dx(-3x + 2y) Right?
```

From here, it's confusing. I know dy/dx can be rewritten as d/dx y, so does that mean that:
Code:
```dy/dx(-3x + 2y) == d/dx y(-3x + 2y)
d/dx y(-3x + 2y) == d/dx(-3xy + 2y^2)
d/dx(-3xy + 2y^2) == d/dx(-3xy) + d/dx(2y^2)
d/dx(-3xy) + d/dx(2y^2) == -3y + d/dx(2y^2)

... and finally:
d/dx f = 10x - 6y + d/dx(2y^2) ???
```

3) Assuming I'm wrong, can you show me the step by step solution to the above problem, please?
09 Nov 2010, 00:16
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Explanations are presented in the picture attached below (PDF version of it you may see here):

09 Nov 2010, 11:35
Tyler

Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
Thanks man. I appreciate the time you must have taken to give such a detailed explanation.

What would make x and y dependant? By dependant, do you mean one is used in the others definition. Like y is defined in terms of x for example?

Should the planes z=10x-3y and z=-3x+2y be tangent to z=5x^2-3xy+y^2? The graph I made with Maxima seems to contradict my assumption that they should be. You can see how z=10x-3y crosses into the bottom of the "dip" in z=5x^2-3xy+y^2 to create a flat place and a slight change in color.

 Description: Filesize: 100.12 KB Viewed: 3134 Time(s)

09 Nov 2010, 22:49
ctl3d32

Joined: 30 Dec 2009
Posts: 204
Location: Brazil
ctl3d32
It is tangent. You see this glitch bacause your curve has too few polygons. Try increasing the number of polygons and you will see it is tangent.
09 Nov 2010, 23:14
Tyler

Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
clt3d32, it is a graph of a function and it's indefinite derivative. I was graphing f(x,y), d/dx f(x,y), and d/dy f(x,y) which aren't tangent. I meant to graph f(x,y) and a definite derivative for a point. I've got a function now to graph d/dx tangents but I'm having trouble with the concept of tangents in y... I post a complete graph when I've got both working(If I do...).
10 Nov 2010, 01:19
Tyler

Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
How 'bout now?

Is there a general form for this? Given df/dx and df/dy.

 Description: Filesize: 99.88 KB Viewed: 3103 Time(s)

10 Nov 2010, 05:26
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Some explanations are presented in the following pictures (printer-friendly PDF document you may see here ):
13 Nov 2010, 00:15
ctl3d32

Joined: 30 Dec 2009
Posts: 204
Location: Brazil
ctl3d32
Nice! You've made me remember some calculus.
13 Nov 2010, 01:59
DarkAlchemist

Joined: 08 Oct 2010
Posts: 108
DarkAlchemist
"I know dy/dx can be rewritten as d/dx y"

Wouldn't that simplify down to x/y since d resolves to 1 so we do not need d at all in that case?
13 Nov 2010, 09:06
Tyler

Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
DarkAlchemist wrote:
"I know dy/dx can be rewritten as d/dx y"

Wouldn't that simplify down to x/y since d resolves to 1 so we do not need d at all in that case?
d is the operator of differentiation, and can't cancel out. It comes from Gottfried Leibniz's notation for derivatives. It is the equivalent to Newton's prime notation, ei f(x) = x^2, f'(x) = 2x.
13 Nov 2010, 14:55
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