次の4つの問題に答えます。 (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。 (3) $xu - yv = 0$, $yu + xv = 1$ から $\frac{\partial u}{\partial x}$, $\frac{\partial v}{\partial x}$ を求めます。 (4) $x = e^u \cos v$, $y = e^u \sin v$, $z = uv$ から $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$ を求めます。

1. 問題の内容

次の4つの問題に答えます。

(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。

(3) $xu - yv = 0$, $yu + xv = 1$ から $\frac{\partial u}{\partial x}$, $\frac{\partial v}{\partial x}$ を求めます。

(4) $x = e^u \cos v$, $y = e^u \sin v$, $z = uv$ から $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$ 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。$f(x, y) = 0$ より、

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

したがって、

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

$f(x, y) = \sin y + e^x - xy^2 = 0$ より、

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

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

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

(2) 陰関数の微分法を使います。$f(x, y, z) = 0$ より、

$\frac{\partial z}{\partial x} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$

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

$f(x, y, z) = x^2 + y^2 + z^2 - 4z = 0$ より、

$\frac{\partial f}{\partial x} = 2x$

$\frac{\partial f}{\partial y} = 2y$

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

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

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

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

$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x} (\frac{\partial z}{\partial y}) = \frac{\partial}{\partial x} (\frac{y}{2 - z}) = \frac{0 \cdot (2-z) - y \cdot (-\frac{\partial z}{\partial x})}{(2-z)^2} = \frac{y \frac{x}{2-z}}{(2-z)^2} = \frac{xy}{(2-z)^3}$

(3) 与えられた方程式を $x$ で偏微分します。

$xu - yv = 0$ より、$u + x \frac{\partial u}{\partial x} - y \frac{\partial v}{\partial x} = 0$

$yu + xv = 1$ より、$y \frac{\partial u}{\partial x} + v + x \frac{\partial v}{\partial x} = 0$

$\begin{cases} x \frac{\partial u}{\partial x} - y \frac{\partial v}{\partial x} = -u \\ y \frac{\partial u}{\partial x} + x \frac{\partial v}{\partial x} = -v \end{cases}$

$\frac{\partial u}{\partial x} = \frac{-ux + vy}{x^2 + y^2}$

$\frac{\partial v}{\partial x} = \frac{-uy - vx}{x^2 + y^2}$

(4)

$x = e^u \cos v$, $y = e^u \sin v$, $z = uv$ から $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$ を求めます。

$z = z(u(x,y), v(x,y))$

$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} = v\frac{\partial u}{\partial x} + u \frac{\partial v}{\partial x}$

$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} = v\frac{\partial u}{\partial y} + u \frac{\partial v}{\partial y}$

$x^2 + y^2 = e^{2u} \cos^2 v + e^{2u} \sin^2 v = e^{2u}$

$u = \frac{1}{2} \ln (x^2 + y^2)$

$\frac{\partial u}{\partial x} = \frac{1}{2} \frac{2x}{x^2 + y^2} = \frac{x}{x^2 + y^2}$

$\frac{\partial u}{\partial y} = \frac{1}{2} \frac{2y}{x^2 + y^2} = \frac{y}{x^2 + y^2}$

$\frac{y}{x} = \frac{e^u \sin v}{e^u \cos v} = \tan v$

$v = \arctan (\frac{y}{x})$

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

$\frac{\partial v}{\partial y} = \frac{1}{1 + (\frac{y}{x})^2} \cdot (\frac{1}{x}) = \frac{x^2}{x^2 + y^2} (\frac{1}{x}) = \frac{x}{x^2 + y^2}$

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

$\frac{\partial z}{\partial y} = v \frac{y}{x^2 + y^2} + u \frac{x}{x^2 + y^2} = \frac{vy + ux}{x^2 + y^2}$

3. 最終的な答え

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

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

(3) $\frac{\partial u}{\partial x} = \frac{-ux + vy}{x^2 + y^2}$, $\frac{\partial v}{\partial x} = \frac{-uy - vx}{x^2 + y^2}$

(4) $\frac{\partial z}{\partial x} = \frac{vx - uy}{x^2 + y^2}$, $\frac{\partial z}{\partial y} = \frac{vy + ux}{x^2 + y^2}$