(1) $yz + zx + xy = 1$ で定まる陰関数 $z = z(x, y)$ について、$(x, y) = (3, 1)$ での $\frac{\partial^2 z}{\partial x^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求めなさい。 (2) $z^x = y^z$ で定まる陰関数 $z = z(x, y)$ について、$(x, y) = (5, 1)$ での $\frac{\partial^2 z}{\partial y^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求めなさい。

1. 問題の内容

(1) $yz + zx + xy = 1$ で定まる陰関数 $z = z(x, y)$ について、$(x, y) = (3, 1)$ での $\frac{\partial^2 z}{\partial x^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求めなさい。

(2) $z^x = y^z$ で定まる陰関数 $z = z(x, y)$ について、$(x, y) = (5, 1)$ での $\frac{\partial^2 z}{\partial y^2}$ と $\frac{\partial^2 z}{\partial x \partial y}$ を求めなさい。

2. 解き方の手順

(1) $yz + zx + xy = 1$ を $x$ で偏微分すると、

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

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

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

$yz + zx + xy = 1$ を $y$ で偏微分すると、

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(2) $z^x = y^z$ の両辺の自然対数をとると、

$x\ln z = z\ln y$

$x\ln z = z\ln y$ を $y$ で偏微分すると、

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

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

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

$x\ln z = z\ln y$ を $x$ で偏微分すると、

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

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

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

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

$\frac{\partial z}{\partial y}(5, 1) = \frac{1^2}{5(1) - 1 \ln 1} = \frac{1}{5 - 0} = \frac{1}{5}$

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

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

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

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

$= \frac{2(1)(\frac{1}{5})(5(1) - 1\ln 1) - (1)^2(5 - \frac{1}{5}\ln 1 - \frac{1}{1})}{(5(1) - 1\ln 1)^2} = \frac{\frac{2}{5}(5) - (5 - 0 - 1)}{25} = \frac{2 - 4}{25} = -\frac{2}{25}$

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

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

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

$= -\frac{(\frac{1}{5}\ln 1 + \frac{1}{5})(5 - 1\ln 1) - 1\ln 1(-\frac{1}{5}\ln 1 - \frac{1}{1})}{(5 - 1\ln 1)^2} = -\frac{(\frac{1}{5})(5) - 0}{25} = -\frac{1}{25}$

3. 最終的な答え

(1) $\frac{\partial^2 z}{\partial x^2}(3, 1) = \frac{1}{16}$, $\frac{\partial^2 z}{\partial x \partial y}(3, 1) = -\frac{1}{16}$

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