(1) 陰関数 $yz + zx + xy = 1$ について、$(x,y) = (3,1)$ での $\frac{\partial^2 z}{\partial x^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求めます。 (2) 陰関数 $z^x = y^z$ について、$(x,y) = (5,1)$ での $\frac{\partial^2 z}{\partial y^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求めます。

1. 問題の内容

(1) 陰関数 $yz + zx + xy = 1$ について、$(x,y) = (3,1)$ での $\frac{\partial^2 z}{\partial x^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求めます。

(2) 陰関数 $z^x = y^z$ について、$(x,y) = (5,1)$ での $\frac{\partial^2 z}{\partial y^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求めます。

2. 解き方の手順

(1) $yz + zx + xy = 1$ の場合

まず、$z$ を $x$ と $y$ の関数として、与えられた式を $x$ で偏微分します。

$y \frac{\partial z}{\partial x} + z + x \frac{\partial z}{\partial x} + y = 0$

$\frac{\partial z}{\partial x} = -\frac{z+y}{x+y}$

次に、$x=3, y=1$ を代入すると、 $z + 3z + 3 = 1$ より $4z = -2$ となり $z = -\frac{1}{2}$。

従って、$\frac{\partial z}{\partial x}(3,1) = -\frac{-\frac{1}{2}+1}{3+1} = -\frac{\frac{1}{2}}{4} = -\frac{1}{8}$

$\frac{\partial z}{\partial x} = -\frac{z+y}{x+y}$ をさらに $x$ で偏微分します。

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

ここに$x=3, y=1, z=-\frac{1}{2}, \frac{\partial z}{\partial x} = -\frac{1}{8}$を代入すると

$\frac{\partial^2 z}{\partial x^2} = -\frac{-\frac{1}{8}(3+1) - (-\frac{1}{2}+1)}{(3+1)^2} = -\frac{-\frac{1}{8}(4) - (\frac{1}{2})}{16} = -\frac{-\frac{1}{2}-\frac{1}{2}}{16} = -\frac{-1}{16} = \frac{1}{16}$

次に、$\frac{\partial z}{\partial x} = -\frac{z+y}{x+y}$ を $y$ で偏微分します。

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

$y \frac{\partial z}{\partial y} + z + x \frac{\partial z}{\partial y} + x = 0$ より、

$\frac{\partial z}{\partial y} = -\frac{z+x}{x+y}$

$x=3, y=1, z = -\frac{1}{2}$ を代入すると、$\frac{\partial z}{\partial y}(3,1) = -\frac{-\frac{1}{2}+3}{3+1} = -\frac{\frac{5}{2}}{4} = -\frac{5}{8}$

$\frac{\partial^2 z}{\partial x \partial y} = -\frac{(-\frac{5}{8}+1)(3+1) - (-\frac{1}{2}+1)}{(3+1)^2} = -\frac{(\frac{3}{8})(4) - (\frac{1}{2})}{16} = -\frac{\frac{3}{2} - \frac{1}{2}}{16} = -\frac{1}{16}$

(2) $z^x = y^z$ の場合

両辺の自然対数を取ると、

$x \ln z = z \ln y$

これを$y$で偏微分すると

$\frac{x}{z} \frac{\partial z}{\partial y} = \frac{z}{y} + \ln y \frac{\partial z}{\partial y}$

$\frac{\partial z}{\partial y} (\frac{x}{z} - \ln y) = \frac{z}{y}$

$\frac{\partial z}{\partial y} = \frac{z/y}{x/z - \ln y} = \frac{z^2}{y(x-z \ln y)}$

$x=5, y=1$を代入すると、$z^5 = 1^z = 1$より$z=1$

$\frac{\partial z}{\partial y} = \frac{1^2}{1(5 - 1 \ln 1)} = \frac{1}{5}$

$\frac{\partial z}{\partial y} = \frac{z^2}{y(x-z \ln y)}$を$y$で偏微分すると

$\frac{\partial^2 z}{\partial y^2} = \frac{2z\frac{\partial z}{\partial y}y(x-z \ln y) - z^2[(x-z \ln y)+y(-\frac{\partial z}{\partial y}\ln y - \frac{z}{y})]}{y^2(x-z \ln y)^2}$

$x=5, y=1, z=1, \frac{\partial z}{\partial y} = \frac{1}{5}$を代入すると

$\frac{\partial^2 z}{\partial y^2} = \frac{2 \cdot \frac{1}{5} \cdot 1 \cdot (5 - 1 \cdot 0) - 1^2 [(5-0) + 1(-\frac{1}{5} \cdot 0 - \frac{1}{1})]}{1^2 (5-0)^2} = \frac{2/5 \cdot 5 - (5-1)}{25} = \frac{2-4}{25} = -\frac{2}{25}$

$x \ln z = z \ln y$を$x$で偏微分すると

$\ln z + \frac{x}{z} \frac{\partial z}{\partial x} = \ln y \frac{\partial z}{\partial x}$

$\frac{\partial z}{\partial x} (\frac{x}{z} - \ln y) = - \ln z$

$\frac{\partial z}{\partial x} = \frac{-\ln z}{\frac{x}{z} - \ln y} = \frac{-z \ln z}{x-z \ln y}$

$x=5, y=1, z=1$を代入すると

$\frac{\partial z}{\partial x} = \frac{-1 \ln 1}{5-1 \ln 1} = \frac{0}{5} = 0$

$\frac{\partial z}{\partial x} = \frac{-z \ln z}{x-z \ln y}$を$y$で偏微分すると

$\frac{\partial^2 z}{\partial x \partial y} = \frac{-(\frac{\partial z}{\partial y} \ln z + z \frac{1}{z} \frac{\partial z}{\partial y}) (x-z \ln y) - (-z \ln z) (- \frac{\partial z}{\partial y} \ln y - \frac{z}{y})}{(x-z \ln y)^2}$

$x=5, y=1, z=1, \frac{\partial z}{\partial y} = \frac{1}{5}$を代入すると

$\frac{\partial^2 z}{\partial x \partial y} = \frac{-(\frac{1}{5} \ln 1 + 1 \cdot \frac{1}{1} \cdot \frac{1}{5}) (5 - 1 \ln 1) - (-1 \ln 1)(-\frac{1}{5}\ln 1 - \frac{1}{1})}{(5-1 \ln 1)^2} = \frac{-(\frac{1}{5})(5) - 0}{25} = \frac{-1}{25}$

3. 最終的な答え

(1) $\frac{\partial^2 z}{\partial x^2} = \frac{1}{16}$, $\frac{\partial^2 z}{\partial x \partial y} = -\frac{1}{16}$

(2) $\frac{\partial^2 z}{\partial y^2} = -\frac{2}{25}$, $\frac{\partial^2 z}{\partial x \partial y} = -\frac{1}{25}$