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

1. 問題の内容

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

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

2. 解き方の手順

(1) $yz + zx + xy = 1$

まず、$z$ を $x$ および $y$ で偏微分する。

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

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

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

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

$(x, y) = (3, 1)$ のとき、 $z + 3z + 3 = 1$ より $4z = -2$ なので、$z = - \frac{1}{2}$

$\frac{\partial z}{\partial x} |_{(3,1)} = - \frac{-1/2 + 1}{3 + 1} = - \frac{1/2}{4} = - \frac{1}{8}$

$\frac{\partial z}{\partial y} |_{(3,1)} = - \frac{-1/2 + 3}{3 + 1} = - \frac{5/2}{4} = - \frac{5}{8}$

次に、$\frac{\partial^2 z}{\partial x^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求める。

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

$\frac{\partial^2 z}{\partial x^2} |_{(3,1)} = - \frac{(-1/8)(3+1) - (-1/2+1)}{(3+1)^2} = - \frac{-1/2 - 1/2}{16} = \frac{1}{16}$

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

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

(2) $z^x = y^z$

両辺の対数をとると、 $x \ln z = z \ln y$

$z$ を $x$ および $y$ で偏微分する。

$\ln z + x \frac{1}{z} \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x} \ln y$

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

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

$x \frac{1}{z} \frac{\partial z}{\partial y} = \frac{\partial z}{\partial y} \ln y + z \frac{1}{y}$

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

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

$(x, y) = (5, 1)$ のとき、$z^5 = 1^z = 1$ より $z = 1$

$\frac{\partial z}{\partial x} |_{(5,1)} = \frac{1 \ln 1}{1 \ln 1 - 5} = \frac{0}{-5} = 0$

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

次に、$\frac{\partial^2 z}{\partial y^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求める。

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

$\frac{\partial^2 z}{\partial y^2} |_{(5,1)} = \frac{2(1)(1/5)(1)(5 - 0) - 1 (5 - 0 + 1(0 - 1))}{1^2 (5 - 0)^2} = \frac{2 - 4}{25} = - \frac{2}{25}$

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

$\frac{\partial^2 z}{\partial x \partial y} |_{(5,1)} = \frac{2(1)(0)(1)(5 - 0) - 1 (1(1 - 0))}{1^2 (5 - 0)^2} = - \frac{1}{25}$

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) $\frac{\partial^2 z}{\partial y^2} = -\frac{2}{25}$

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