関数 $z$ が与えられたとき、以下の2階の偏導関数を求める問題です。 * $z_{xx} = \frac{\partial^2 z}{\partial x^2}$ * $z_{yy} = \frac{\partial^2 z}{\partial y^2}$ * $z_{xy} = \frac{\partial^2 z}{\partial x \partial y}$ * $z_{yx} = \frac{\partial^2 z}{\partial y \partial x}$

はい、承知いたしました。問題3の関数 z の 2 階の偏導関数を求める問題ですね。

今回は(1) $z = ax^2 - bxy + cy^2$、(2) $z = \frac{1}{x} - \frac{1}{y}$、(3) $z = \sin(ax + by)$の3つの問題について解答します。

1. 問題の内容

関数 $z$ が与えられたとき、以下の2階の偏導関数を求める問題です。

* $z_{xx} = \frac{\partial^2 z}{\partial x^2}$

* $z_{yy} = \frac{\partial^2 z}{\partial y^2}$

* $z_{xy} = \frac{\partial^2 z}{\partial x \partial y}$

* $z_{yx} = \frac{\partial^2 z}{\partial y \partial x}$

2. 解き方の手順

**3.(1) $z = ax^2 - bxy + cy^2$**

* $z_x = \frac{\partial z}{\partial x} = 2ax - by$

* $z_y = \frac{\partial z}{\partial y} = -bx + 2cy$

次に、2階偏導関数を計算します。

* $z_{xx} = \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(2ax - by) = 2a$

* $z_{yy} = \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(-bx + 2cy) = 2c$

* $z_{xy} = \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x}(-bx + 2cy) = -b$

* $z_{yx} = \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y}(2ax - by) = -b$

**3.(2) $z = \frac{1}{x} - \frac{1}{y}$**

* $z_x = \frac{\partial z}{\partial x} = -\frac{1}{x^2}$

* $z_y = \frac{\partial z}{\partial y} = \frac{1}{y^2}$

次に、2階偏導関数を計算します。

* $z_{xx} = \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(-\frac{1}{x^2}) = \frac{2}{x^3}$

* $z_{yy} = \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(\frac{1}{y^2}) = -\frac{2}{y^3}$

* $z_{xy} = \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x}(\frac{1}{y^2}) = 0$

* $z_{yx} = \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y}(-\frac{1}{x^2}) = 0$

**3.(3) $z = \sin(ax + by)$**

* $z_x = \frac{\partial z}{\partial x} = a\cos(ax + by)$

* $z_y = \frac{\partial z}{\partial y} = b\cos(ax + by)$

次に、2階偏導関数を計算します。

* $z_{xx} = \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(a\cos(ax + by)) = -a^2\sin(ax + by)$

* $z_{yy} = \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(b\cos(ax + by)) = -b^2\sin(ax + by)$

* $z_{xy} = \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x}(b\cos(ax + by)) = -ab\sin(ax + by)$

* $z_{yx} = \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y}(a\cos(ax + by)) = -ab\sin(ax + by)$

3. 最終的な答え

**3.(1)**

* $z_{xx} = 2a$

* $z_{yy} = 2c$

* $z_{xy} = -b$

* $z_{yx} = -b$

**3.(2)**

* $z_{xx} = \frac{2}{x^3}$

* $z_{yy} = -\frac{2}{y^3}$

* $z_{xy} = 0$

* $z_{yx} = 0$

**3.(3)**

* $z_{xx} = -a^2\sin(ax + by)$

* $z_{yy} = -b^2\sin(ax + by)$

* $z_{xy} = -ab\sin(ax + by)$

* $z_{yx} = -ab\sin(ax + by)$