6.8 陰関数

1.

(a) $f(x,y) = x - y^{2} - 1 = 0$とおき，全微分を求めると， $df = f_{x}dx + f_{y}dy = dx - 2ydy = 0$．これより， $\frac{dy}{dx} = \frac{1}{2y}$. 次に， $\frac{d^{2}y}{dx^{2}}$を求める．

$$\frac{d^{2}y}{dx^{2}} = \frac{d(\frac{dy}{dx})}{dy} \frac{dy}{dx} = \frac{-2}{4y^{2}}\cdot \frac{1}{2y} = -\frac{1}{8y^{3}}$$

(b) $f(x,y) = x^{2} + xy + 2y^{2} - 1 = 0$とおき，全微分を求めると， $df = f_{x}dx + f_{y}dy = (2x+y)dx + (x + 4y)dy = 0$．これより， $\frac{dy}{dx} = -\frac{2x+y}{x + 4y}$. 次に， $\frac{d^{2}y}{dx^{2}}$を求める．

$$\frac{d^{2}y}{dx^{2}} = \frac{-(2 + \frac{dy}{dx})(x+4y) + (2x+y)(1 + 4\frac{dy}{dx})}{(x+4y)^{2}}$$

$$= \frac{-2(x+4y) + 2x+y - (x+4y)(\frac{-(2x+y)}{x+4y}) + (2x+y)(\frac{-4(2x+y)}{x+4y})}{(x+4y)^{2}}$$

$$= \frac{-7y(x+4y) + (x+4y)(2x+y) - 4(2x+y)^{2}}{(x+4y)^{3}}$$

$$= \frac{-7xy - 28y^{2} + 2x^{2} + 9xy + 4y^{2} - 16x^{2} - 16xy - 4y^{2}}{(x+4y)^{3}}$$

$$= \frac{-14x^{2} - 14xy - 28y^{2}}{(x+4y)^{3}} = -\frac{14(x^{2} + xy + 2y^{2}}{(x+4y)^{3}}$$

(c) $f(x,y) = x-e^{y} = 0$とおき，全微分を求めると， $df = f_{x}dx + f_{y}dy = dx + -e^{y}dy = 0$．これより， $\frac{dy}{dx} = \frac{1}{e^{y}} = e^{-y}$. 次に， $\frac{d^{2}y}{dx^{2}}$を求める．

$$\frac{d^{2}y}{dx^{2}} = e^{-y}\frac{dy}{dx} = e^{-y}\cdot e^{-y} = e^{-2y}$$

(d) $f(x,y) = x^{3} - 3xy + y^{3} - 1 = 0$とおき，全微分を求めると， $df = f_{x}dx + f_{y}dy = (3x^{2} - 3y)dx + (-3x + 3y^{2})dy = 0$．これより， $\frac{dy}{dx} = \frac{3x^{2} - 3y}{3x - 3y^{2}} = \frac{x^{2} - y}{x- y^{2}}$. 次に， $\frac{d^{2}y}{dx^{2}}$を求める．

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

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

$$= \frac{(x-y^{2})\left(2x(x-y^{2}) - 2(x^{2} - y)\right) + 2y(x^{2} - y)^2}{(x- y^{2})^{3}}$$

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

$$= \frac{-2x^2 y^2 + 2xy^4 + 2xy - 2y^3 + 2x^4 y - 4x^2 y^2 + 2y^3}{(x - y^2)^{3}}$$

$$= \frac{2x^4 y - 6x^2 y^2 + 2xy^4 + 2xy}{(x - y^2)^{3}}$$

$$= \frac{2xy(x^{3} - 3xy + y^{3} + 1)}{(x-y^{2})^{3}}$$

2.

(a) $f(x,y,z) = x^{2} + y^{2} + z^2 - 4 = 0$
$g(x,y,z) = x + y + z - 1 = 0$の全微分を求めると，

$$\begin{array}{l} df = 2xdx + 2ydy + 2zdz = 0\\ dg = dx + dy + dz = 0 \end{array}$$

$dz = -dx -dy$で置き換えると，

$$2xdx + 2y dy + 2z(-dx -dy) = 0$$

これより，

$$(2x - 2z)dx + (2y - 2z)dy = 0$$

となり，

$$\frac{dy}{dx} = \frac{-(2x - 2z)}{2y - 2z} = -\frac{z-x}{y-z}$$

次に， $dy = -dz -dx$で置き換えると，

$$2xdx + 2y(-dz - dx) + 2zdz = 0$$

これより，

$$(2x - 2y)dx + (2z - 2y)dz = 0$$

となり，

$$\frac{dz}{dx} = \frac{2x - 2y}{2y - 2z} = \frac{x-y}{y-z}$$

(b) $f(x,y,z) = xyz - 1 = 0$
$g(x,y,z) = x + y + z - 1 = 0$の全微分を求めると，

$$\begin{array}{l} df = yzdx + xzdy + xydz = 0\\ dg = dx + dy + dz = 0 \end{array}$$

$dz = -dx -dy$で置き換えると，

$$yzdx + xzdy + xy(-dx -dy) = 0$$

これより，

$$(yz - xy)dx + (xz - xy)dy = 0$$

となり，

$$\frac{dy}{dx} = \frac{y(z-x)}{x(y-z)}$$

次に， $dy = -dz -dx$で置き換えると，

$$yzdx + xz(-dz - dx) + xydz = 0$$

これより，

$$(yz - xz)dx + (xy - xz)dz = 0$$

となり，

$$\frac{dz}{dx} = \frac{z(x-y)}{x(y-z)}$$