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6の解答を求めてみる。

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

u = 2 x + 7 y

とおくと、

\begin{array}{l} 2 \frac{\partial g}{\partial y} = 2 \frac{\partial g}{\partial u} 7 = 14 \frac{\partial g}{\partial u} \\ 7 \frac{\partial g}{\partial x} = 7 \frac{\partial g}{\partial u} 2 = 14 \frac{\partial g}{\partial u} \end{array}

よって、

2 \frac{\partial g}{\partial y} = 7 \frac{\partial g}{\partial x}

u = 2 x^{3} + 3 y^{2}

とおくと、

\begin{array}{l} y \frac{\partial g}{\partial x} = y \frac{\partial f}{\partial u} 6 x^{2} = 6 x^{2} y \frac{\partial f}{\partial u} \\ x^{2} \frac{\partial g}{\partial y} = x^{2} \frac{\partial f}{\partial u} 6 y = 6 x^{2} y \frac{\partial f}{\partial u} \end{array}

よって、

y \frac{\partial g}{\partial x} = x^{2} \frac{\partial g}{\partial y}