次の関数 $z$ の偏導関数 $\frac{\partial z}{\partial x}$ と $\frac{\partial z}{\partial y}$ を求めます。 (1) $z = x^3 + 2xy^2 + 3y^3$ (2) $z = \frac{xy}{x-y}$ (3) $z = x^2\sqrt{1-y^2}$ (4) $z = \cos(x^2 + xy)$ (5) $z = xye^{xy}$ (6) $z = x\log(x+y)$ (7) $z = \arcsin(x^2y)$ (8) $z = \arctan(\frac{y}{x})$

はい、承知いたしました。与えられた関数の偏導関数を求めます。

1. 問題の内容

次の関数 $z$ の偏導関数 $\frac{\partial z}{\partial x}$ と $\frac{\partial z}{\partial y}$ を求めます。

(1) $z = x^3 + 2xy^2 + 3y^3$

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

(3) $z = x^2\sqrt{1-y^2}$

(4) $z = \cos(x^2 + xy)$

(5) $z = xye^{xy}$

(6) $z = x\log(x+y)$

(7) $z = \arcsin(x^2y)$

(8) $z = \arctan(\frac{y}{x})$

2. 解き方の手順

各関数について、$x$ と $y$ でそれぞれ偏微分を行います。

(1) $z = x^3 + 2xy^2 + 3y^3$

$\frac{\partial z}{\partial x} = 3x^2 + 2y^2$

$\frac{\partial z}{\partial y} = 4xy + 9y^2$

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

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

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

(3) $z = x^2\sqrt{1-y^2}$

$\frac{\partial z}{\partial x} = 2x\sqrt{1-y^2}$

$\frac{\partial z}{\partial y} = x^2\frac{-2y}{2\sqrt{1-y^2}} = \frac{-x^2y}{\sqrt{1-y^2}}$

(4) $z = \cos(x^2 + xy)$

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

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

(5) $z = xye^{xy}$

$\frac{\partial z}{\partial x} = ye^{xy} + xy^2e^{xy} = ye^{xy}(1+xy)$

$\frac{\partial z}{\partial y} = xe^{xy} + x^2ye^{xy} = xe^{xy}(1+xy)$

(6) $z = x\log(x+y)$

$\frac{\partial z}{\partial x} = \log(x+y) + x\frac{1}{x+y} = \log(x+y) + \frac{x}{x+y}$

$\frac{\partial z}{\partial y} = x\frac{1}{x+y} = \frac{x}{x+y}$

(7) $z = \arcsin(x^2y)$

$\frac{\partial z}{\partial x} = \frac{2xy}{\sqrt{1-(x^2y)^2}} = \frac{2xy}{\sqrt{1-x^4y^2}}$

$\frac{\partial z}{\partial y} = \frac{x^2}{\sqrt{1-(x^2y)^2}} = \frac{x^2}{\sqrt{1-x^4y^2}}$

(8) $z = \arctan(\frac{y}{x})$

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

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

3. 最終的な答え

(1) $\frac{\partial z}{\partial x} = 3x^2 + 2y^2$, $\frac{\partial z}{\partial y} = 4xy + 9y^2$

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

(3) $\frac{\partial z}{\partial x} = 2x\sqrt{1-y^2}$, $\frac{\partial z}{\partial y} = \frac{-x^2y}{\sqrt{1-y^2}}$

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

(5) $\frac{\partial z}{\partial x} = ye^{xy}(1+xy)$, $\frac{\partial z}{\partial y} = xe^{xy}(1+xy)$

(6) $\frac{\partial z}{\partial x} = \log(x+y) + \frac{x}{x+y}$, $\frac{\partial z}{\partial y} = \frac{x}{x+y}$

(7) $\frac{\partial z}{\partial x} = \frac{2xy}{\sqrt{1-x^4y^2}}$, $\frac{\partial z}{\partial y} = \frac{x^2}{\sqrt{1-x^4y^2}}$

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