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I have this little script to derive implicitly

$PrePrint = # /. {D[y_, x_, NonConstants -> {y_}] :> y'[x]} &;

Example:

D[x == y^3 + x y, x, NonConstants -> y]

1 == y + x y'[x] + 3 y^2*y'[x]

and then, to obtain the expression for y'[x], I use

Solve[1 == y + x y'[x] + 3 y^2*y'[x],y'[x]] //FullSimplify

How can I get all these steps in one line?

I'd like to have something like

$PrePrint = # /. {D[y_, x_, NonConstants -> {y_}] :> y'[x]} &; //Solve...
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  • $\begingroup$ I don't see a question anywhere… $\endgroup$ Jul 6, 2015 at 0:02
  • $\begingroup$ Look up Dt[]. $\endgroup$ Jul 6, 2015 at 0:56
  • $\begingroup$ If I saw that, but I want to publish online with the Solve not like having to go to take the result of the derivative and clearance and y '[x] $\endgroup$ Jul 6, 2015 at 1:02
  • 2
    $\begingroup$ I cannot understand the question at all. You may need a better english translation before posting here. $\endgroup$
    – Jens
    Jul 6, 2015 at 1:19
  • $\begingroup$ Related: (1945), (24422), (52284) $\endgroup$
    – Michael E2
    Jul 6, 2015 at 4:23

1 Answer 1

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If my interpretation of your question is correct, the following code should produce the desired behaviour.

prePrint[input_] := 
  Module[{solveFor}, 
   input /. {D[y_, x_, NonConstants -> {y_}] :> (solveFor = y'[x])} //
      If[OwnValues[solveFor] === {}, input, Solve[#, solveFor]] & // 
    FullSimplify];
$PrePrint = prePrint;

Test

D[x == y^3 + x y, x, NonConstants -> y]

{{y'[x]] -> (1 - y)/(x + 3 y^2)}}

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1
  • $\begingroup$ if just that is, thank you, sorry for my English $\endgroup$ Jul 6, 2015 at 2:02

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