$z = f(x, y)$ であり、$x = u + v$、$y = uv$ であるとき、以下の偏微分を計算し、空欄を埋める問題です。 1. $z_u = z_x + \boxed{①} z_y$, $z_v = z_x + \boxed{②} z_y$

1. 問題の内容

$z = f(x, y)$ であり、$x = u + v$、$y = uv$ であるとき、以下の偏微分を計算し、空欄を埋める問題です。

1. $z_u = z_x + \boxed{①} z_y$, $z_v = z_x + \boxed{②} z_y$

2. $z_{uu} - 2z_{uv} + z_{vv} = (x^2 - \boxed{③}y) z_{yy} - \boxed{④} z_y$

2. 解き方の手順

1. まず、$z_u$ と $z_v$ を計算します。

偏微分の連鎖律より、

$z_u = \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}$

$z_v = \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}$

ここで、$x = u + v$、$y = uv$ より、

$\frac{\partial x}{\partial u} = 1$, $\frac{\partial y}{\partial u} = v$

$\frac{\partial x}{\partial v} = 1$, $\frac{\partial y}{\partial v} = u$

よって、

$z_u = z_x (1) + z_y (v) = z_x + v z_y$

$z_v = z_x (1) + z_y (u) = z_x + u z_y$

したがって、$\boxed{①} = v$、$\boxed{②} = u$ となります。

2. 次に、$z_{uu} - 2z_{uv} + z_{vv}$ を計算します。

まず、$z_u = z_x + v z_y$ と $z_v = z_x + u z_y$ を用いて、$z_{uu}, z_{uv}, z_{vv}$ を計算します。

$z_{uu} = \frac{\partial z_u}{\partial u} = \frac{\partial}{\partial u} (z_x + v z_y) = \frac{\partial z_x}{\partial u} + v \frac{\partial z_y}{\partial u}$

$z_{uv} = \frac{\partial z_u}{\partial v} = \frac{\partial}{\partial v} (z_x + v z_y) = \frac{\partial z_x}{\partial v} + \frac{\partial}{\partial v} (v z_y) = \frac{\partial z_x}{\partial v} + z_y + v \frac{\partial z_y}{\partial v}$

$z_{vv} = \frac{\partial z_v}{\partial v} = \frac{\partial}{\partial v} (z_x + u z_y) = \frac{\partial z_x}{\partial v} + u \frac{\partial z_y}{\partial v}$

ここで、$z_x$ と $z_y$ について、再び連鎖律を適用すると、

$\frac{\partial z_x}{\partial u} = \frac{\partial z_x}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z_x}{\partial y} \frac{\partial y}{\partial u} = z_{xx}(1) + z_{xy}(v)$

$\frac{\partial z_x}{\partial v} = \frac{\partial z_x}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z_x}{\partial y} \frac{\partial y}{\partial v} = z_{xx}(1) + z_{xy}(u)$

$\frac{\partial z_y}{\partial u} = \frac{\partial z_y}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z_y}{\partial y} \frac{\partial y}{\partial u} = z_{yx}(1) + z_{yy}(v)$

$\frac{\partial z_y}{\partial v} = \frac{\partial z_y}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z_y}{\partial y} \frac{\partial y}{\partial v} = z_{yx}(1) + z_{yy}(u)$

よって、

$z_{uu} = z_{xx} + v z_{xy} + v (z_{yx} + v z_{yy}) = z_{xx} + 2v z_{xy} + v^2 z_{yy}$

$z_{uv} = z_{xx} + u z_{xy} + z_y + v (z_{yx} + u z_{yy}) = z_{xx} + (u+v) z_{xy} + z_y + uv z_{yy}$

$z_{vv} = z_{xx} + u z_{xy} + u (z_{yx} + u z_{yy}) = z_{xx} + 2u z_{xy} + u^2 z_{yy}$

$z_{uu} - 2z_{uv} + z_{vv} = (z_{xx} + 2v z_{xy} + v^2 z_{yy}) - 2(z_{xx} + (u+v) z_{xy} + z_y + uv z_{yy}) + (z_{xx} + 2u z_{xy} + u^2 z_{yy})$

$= z_{xx} (1 - 2 + 1) + z_{xy}(2v - 2u - 2v + 2u) + z_{yy} (v^2 - 2uv + u^2) - 2z_y$

$= (u-v)^2 z_{yy} - 2 z_y = (x^2 - 4y) z_{yy} - 2 z_y$

したがって、$\boxed{③} = 4$、$\boxed{④} = 2$ となります。

3. 最終的な答え

①: 4

②: 2

1. $z_u = z_x + v z_y$, $z_v = z_x + u z_y$

2. $z_{uu} - 2z_{uv} + z_{vv} = (x^2 - 4y) z_{yy} - 2 z_y$