(1) $f(x, y) = \sin y + e^x - xy^2 = 0$ から $\frac{dy}{dx}$ を求めます。 (2) $f(x, y, z) = x^2 + y^2 + z^2 - 4z = 0$ から $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$, $\frac{\partial^2 z}{\partial x^2}$, $\frac{\partial^2 z}{\partial x \partial y}$ を求めます。

1. 問題の内容

(1) $f(x, y) = \sin y + e^x - xy^2 = 0$ から $\frac{dy}{dx}$ を求めます。

(2) $f(x, y, z) = x^2 + y^2 + z^2 - 4z = 0$ から $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$, $\frac{\partial^2 z}{\partial x^2}$, $\frac{\partial^2 z}{\partial x \partial y}$ を求めます。

2. 解き方の手順

(1) 陰関数の微分法を使います。$f(x, y) = 0$ を $x$ で微分すると、

$\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx} = 0$

より、

$\frac{dy}{dx} = - \frac{\partial f / \partial x}{\partial f / \partial y}$

となります。

$\frac{\partial f}{\partial x} = e^x - y^2$

$\frac{\partial f}{\partial y} = \cos y - 2xy$

したがって、

$\frac{dy}{dx} = - \frac{e^x - y^2}{\cos y - 2xy} = \frac{y^2 - e^x}{\cos y - 2xy}$

(2) まず、$f(x, y, z) = x^2 + y^2 + z^2 - 4z = 0$を$x$と$y$で偏微分します。

$\frac{\partial f}{\partial x} = 2x + 2z \frac{\partial z}{\partial x} - 4\frac{\partial z}{\partial x} = 0$

$(2z - 4) \frac{\partial z}{\partial x} = -2x$

$\frac{\partial z}{\partial x} = \frac{-2x}{2z - 4} = \frac{x}{2 - z}$

$\frac{\partial f}{\partial y} = 2y + 2z \frac{\partial z}{\partial y} - 4\frac{\partial z}{\partial y} = 0$

$(2z - 4) \frac{\partial z}{\partial y} = -2y$

$\frac{\partial z}{\partial y} = \frac{-2y}{2z - 4} = \frac{y}{2 - z}$

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

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

$\frac{\partial z}{\partial x} = \frac{x}{2 - z}$を代入すると、

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

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

$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y} \left( \frac{x}{2 - z} \right) = \frac{0 \cdot (2 - z) - x \cdot (-\frac{\partial z}{\partial y})}{(2 - z)^2} = \frac{x \frac{\partial z}{\partial y}}{(2 - z)^2}$

$\frac{\partial z}{\partial y} = \frac{y}{2 - z}$を代入すると、

$\frac{\partial^2 z}{\partial x \partial y} = \frac{x \cdot \frac{y}{2 - z}}{(2 - z)^2} = \frac{xy}{(2 - z)^3}$

3. 最終的な答え

(1) $\frac{dy}{dx} = \frac{y^2 - e^x}{\cos y - 2xy}$

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

$\frac{\partial z}{\partial y} = \frac{y}{2 - z}$

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

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