Re: Implicit differentiation
- To: mathgroup@smc.vnet.net
- Subject: [mg11399] Re: [mg11372] Implicit differentiation
- From: Bob Hanlon <BobHanlon@aol.com>
- Date: Sun, 8 Mar 1998 20:13:21 -0500
eqn = x^3 + y^3 == 2;
soln = Solve[eqn, y][[1]];
NestList[Dt[#1, x] & , soln, 3] // Simplify
\!\(\*FormBox[
RowBox[{"{",
RowBox[{\({y \[Rule] \@\(2 - x\^3\)\%3}\), ",",
RowBox[{"{",
RowBox[{
FractionBox[\(\[DifferentialD]y\), \(\[DifferentialD]x\),
MultilineFunction->None], "\[Rule]",
\(-\(x\^2\/\((2 - x\^3)\)\^\(2/3\)\)\)}], "}"}], ",",
RowBox[{"{",
RowBox[{
FractionBox[\(\[DifferentialD]\^2 y\),
\(\[DifferentialD]x\^2\),
MultilineFunction->None], "\[Rule]",
\(-\(\(4\ x\)\/\((2 - x\^3)\)\^\(5/3\)\)\)}], "}"}], ",",
RowBox[{"{",
RowBox[{
FractionBox[\(\[DifferentialD]\^3 y\),
\(\[DifferentialD]x\^3\),
MultilineFunction->None], "\[Rule]",
\(-\(\(8\ \((2\ x\^3 + 1)\)\)\/\((2 - x\^3)\)\^\(8/3\)\)\)}],
"}"}]}], "}"}], TraditionalForm]\)
Bob Hanlon
In a message dated 3/7/98 4:26:26 AM, mavalosjr@aol.com wrote:
>Any ideas how I can plug into mathematica to find the 2nd or third
>derivative of an implicit function? The first derivative is okay. I'm
>using( Example): in:eq= x^3 + y^3 ==2
> in: step1= Dt[eq,x]
> in: step2=Solve[step1, Dt[y,x] (* This
>gives me the first derivative*). Thanks again for any suggestions.
>Mavalosjr@aol.com
>Manuel Avalos