Question

用mathematica求偏导

已知

求
的一二阶偏导数。

如何用mathematica计算求解？谢谢。





Answer 1

D[f[p[x, y], w[x, y]], x, y]

Out[2]=
\!\(\*SuperscriptBox[\(w\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y] (
\!\(\*SuperscriptBox[\(w\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y]
\!\(\*SuperscriptBox[\(f\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "2"}], ")"}],
Derivative],
MultilineFunction->None]\)[p[x, y], w[x, y]] +
\!\(\*SuperscriptBox[\(p\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y]
\!\(\*SuperscriptBox[\(f\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[p[x, y], w[x, y]]) +
\!\(\*SuperscriptBox[\(f\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[p[x, y], w[x, y]]
\!\(\*SuperscriptBox[\(p\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y] +
\!\(\*SuperscriptBox[\(f\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[p[x, y], w[x, y]]
\!\(\*SuperscriptBox[\(w\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y] +
\!\(\*SuperscriptBox[\(p\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y] (
\!\(\*SuperscriptBox[\(w\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y]
\!\(\*SuperscriptBox[\(f\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[p[x, y], w[x, y]] +
\!\(\*SuperscriptBox[\(p\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y]
\!\(\*SuperscriptBox[\(f\),
TagBox[
RowBox[{"(",
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[p[x, y], w[x, y]])

追问

请问结果怎么看，没太懂

比如对x一阶偏导应该是

直接得出的结果太复杂没看懂……

非常感谢

追答

你复制到Mathematica中就看到清了。
D[f[p[x, y], w[x, y]], x]

\!\(\*SuperscriptBox[\(f\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[p[x, y], w[x, y]]
\!\(\*SuperscriptBox[\(p\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y] +
\!\(\*SuperscriptBox[\(f\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[p[x, y], w[x, y]]
\!\(\*SuperscriptBox[\(w\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, y]

追问

就是运行了看不懂才问的……

比如 

是什么意思，麻烦稍微解释下……谢谢

追答

函数改名是为了方便输入p,w
f^(0,1)对第2变量求导
p^(1，0)对第1变量求导

Answer 2

a计算求解？谢

追问

能不能详细一点，没看懂。