合成微分律と勾配ベクトルさらに偏微分の計算について 2階微分、2つの関数、等式の証明

学習環境

続解析入門 (原書第2版) (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第4章(合成微分律と勾配ベクトル)、5(さらに偏微分の計算について)の練習問題11、12の解答を求めてみる。

\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} - \frac{\partial f}{\partial v}

\begin{array}{l} \frac{\partial^{2} z}{\partial x \partial y} = (\frac{\partial^{2} f}{\partial u^{2}} - \frac{\partial^{2} f}{\partial u \partial v}) + (\frac{\partial^{2} f}{\partial u \partial v} - \frac{\partial^{2} f}{\partial v^{2}}) \\ = \frac{\partial^{2} f}{\partial u^{2}} - \frac{\partial^{2} f}{\partial v^{2}} \end{array}

\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} - \frac{\partial g}{\partial v}

\frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} f}{\partial u^{2}} - \frac{\partial^{2} g}{\partial v^{2}}

\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} + \frac{\partial g}{\partial v}

\frac{\partial^{2} z}{\partial y^{2}} = \frac{\partial^{2} f}{\partial u^{2}} + \frac{\partial^{2} g}{\partial v^{2}} (- 1) = \frac{\partial^{2} f}{\partial u^{2}} - \frac{\partial^{2} g}{\partial v^{2}}

よって、

\frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} z}{\partial y^{2}} = f'' (u) - g'' (v)