Further expanding d[xy]/dx in the output of a Dt calculation

[John Accardi]

Hello ... any help appreciated ....

Ref:

http://www.accardi.com/ImplicitDifferentiationForumQuery.nb

As you see in the *.nb cited above, during the implicit differentiation
calculation, Mathematica includes d[xy]/dx in the solution. I would
like the product differentiation rule invoked for a complete
differentiation during the Dt calculation. Then, I would hope to see
d[xy]/dx not show in the value of dy/dx in the Solve output.

Here it is in ASCII, but it might be easier to see if you run the
notebook ....

eq5 = x^3-x^2y+2xy^2 == 12
Dt[eq5,x]
-3x^2 dy/dx +3x^2 + 4xy dxy/dx -6xy = 0
Solve[%%,dy/dx]
{{dy/dx -> (3x^2 + 4xy dxy/dx -6xy)/3x^2}}

How can I get Mathematica to show a more complete differentiation by
expanding dxy/dx?

Thank you
Gianni

[telefunkenvf14]

Mathematica is treating 'xy' as a unique variable, not a product.
Include a space between the variables OR use * to indicate
multiplication in your 'eq5' and I think you'll get what you want.

FYI, I looked at your notebook, and noticed you were using traditional
form input and output. Everyone has their own prefs here, but my
experience is that using StandardForm input/output helps you catch
your mistakes, making programming much less painful!! (You can always
make things pretty later...)

In[1]:= eq5 = x^3 - 3 x^2 y + 2 x y^2 == 12
Dt[eq5, x]
Solve[%, Dt[y, x]]

Out[1]= x^3 - 3 x^2 y + 2 x y^2 == 12
Out[2]= 3 x^2 - 6 x y + 2 y^2 - 3 x^2 Dt[y, x] + 4 x y Dt[y, x] == 0
Out[3]= {{Dt[y, x] -> (3 x^2 - 6 x y + 2 y^2)/(x (3 x - 4 y))}}

-RG

[Peter Breitfeld]

John Accardi wrote:

It seems to me, that you missed a space between x and y. xy is not x*y
but a new variable

//Peter
--
_________________________________________________________________
Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de