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偏導関数、内積

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2の解答を求めてみる。

\frac{\partial f}{\partial x} = 3 x^{2} + 3 y z

\frac{\partial f}{\partial y} = 3 x z - 2 y z

\begin{array}{l} \frac{\partial f}{\partial s} = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) \cdot (1, - 1, 2 s) \\ = 3 x^{2} + 3 y z - 3 x z + 2 y z + (3 x y - y^{2}) 2 s \end{array}

\begin{array}{l} \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \frac{x + y}{1 - x y} \\ = \frac{(1 - x y) + (x + y) y}{{(1 - x y)}^{2}} \\ = \frac{1 + y^{2}}{{(1 - x y)}^{2}} \end{array}

\frac{\partial f}{\partial y} = \frac{1 + x^{2}}{{(1 - x y)}^{2}}

\begin{array}{l} \frac{\partial f}{\partial s} = g r a d f (x, y) \cdot (0, \sin (3 t - s)) \\ = \frac{(1 + x^{2}) \sin (3 t - s)}{{(1 - x y)}^{2}} \end{array}

\begin{array}{l} \frac{\partial f}{\partial t} = g r a d f (x, y) \cdot (2 \cos 2 t, - 3 \sin (3 t - s)) \\ = \frac{(1 + y^{2}) 2 \cos 2 t - (1 + x^{2}) 3 \sin (3 t - s)}{{(1 - x y)}^{2}} \end{array}

コード(Wolfram Language, Jupyter)

Plot3D[(x+y)/(1-x y), {x, -5, 5}, {y, -5, 5}]

Plot3D[(Sin[2t] + Cos[3t-s])/(1-Sin[2t]Cos[3t-s]), {s, -5, 5}, {t, -5, 5}]