How can I differentiate a function with respect to several variables and evaluate it at the same time ? I want to specify also the variable index that I want to differentiate and the number of times I do it for each one.
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Is this what you mean?
f[x_, y_] := Sin[x y^2]
Derivative[1, 0][f][x, y]
This is to get the derivative to x.
To get, say, the second derivative to x and the first derivative to y, and evaluate in (0,Pi), you would write
Derivative[2, 1][f][0, Pi]
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2$\begingroup$ You could of course also use
D, but it is a bit less compact:D[f[x, y], {x, 2}, y] /. {x -> 0, y -> Pi}. $\endgroup$ – freddieknets Jun 11 '12 at 14:47 -
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$\begingroup$ Ps Faysal, can you mark the answer as a solution? Thanks! $\endgroup$ – freddieknets Sep 12 '12 at 9:37
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Here's a way a way that avoids the syntax a bit complicated of Derivative when a function has a lot of variables.
SetAttributes[MultiD, HoldFirst];
MultiD[f_[params__]] := f[params];
MultiD[f_[params__], diffVars__] :=
Module[{diffList},
diffList = ConstantArray[0, Length@{params}];
MapThread[(diffList[[#1]] = #2) &, Transpose@{diffVars}];
Derivative[##][f][params]& @@ diffList
];
Example
f[x_, y_] := x^4 y^2;
MultiD[f[1, 2], {2, 1}, {1, 3}]
A similar function can be done to get a derivative as a pure function. I wonder if there's a way to know the number of variables of a pure or interpolated function ?
RemoveHead[h_[args___]] := {args};
NKeys[symbol_] := RemoveHead @@@ DownValues[symbol][[All,1]];
(*For pure or interpolated functions NumberOfVariables needs to be given as option*)
Options[PureD]={"NumberOfVariables"->Automatic};
PureD[f_,diffVars__List,OptionsPattern[]] :=
Module[ {nVars,diffList},
If[ (nVars = OptionValue["NumberOfVariables"])==Automatic,
(*Gets the number of parameters of the last DownValue of f*)
nVars = Length[NKeys[f][[-1]]];
];
diffList = ConstantArray[0,nVars];
MapThread[(diffList[[#1]] = #2)&,Transpose[{diffVars}]];
Derivative[##][f]& @@ diffList
];
Example
PureD[f, {2, 1}, {1, 3}]
SeriesCoefficient[]quickly yields a scaled version of the derivative; figuring out what the scale factor is is left as an exercise. $\endgroup$ – J. M. will be back soon♦ Jun 11 '12 at 17:54