与えられた4つの関数 $z$ をそれぞれ $x$ と $y$ で偏微分する問題です。

1. 問題の内容

与えられた4つの関数 $z$ をそれぞれ $x$ と $y$ で偏微分する問題です。

2. 解き方の手順

(1) $z = \sin(x^2 + y^2)$

* $x$ で偏微分：$\frac{\partial z}{\partial x} = \cos(x^2 + y^2) \cdot 2x = 2x\cos(x^2 + y^2)$

* $y$ で偏微分：$\frac{\partial z}{\partial y} = \cos(x^2 + y^2) \cdot 2y = 2y\cos(x^2 + y^2)$

(2) $z = \arcsin(xy)$

* $x$ で偏微分：$\frac{\partial z}{\partial x} = \frac{1}{\sqrt{1 - (xy)^2}} \cdot y = \frac{y}{\sqrt{1 - x^2y^2}}$

* $y$ で偏微分：$\frac{\partial z}{\partial y} = \frac{1}{\sqrt{1 - (xy)^2}} \cdot x = \frac{x}{\sqrt{1 - x^2y^2}}$

(3) $z = e^{xy}\arctan(y)$

* $x$ で偏微分：$\frac{\partial z}{\partial x} = e^{xy} \cdot y \cdot \arctan(y) = ye^{xy}\arctan(y)$

* $y$ で偏微分：$\frac{\partial z}{\partial y} = e^{xy} \cdot x \cdot \arctan(y) + e^{xy} \cdot \frac{1}{1+y^2} = xe^{xy}\arctan(y) + \frac{e^{xy}}{1+y^2}$

(4) $z = xy\log(2x + y)$

* $x$ で偏微分：$\frac{\partial z}{\partial x} = y\log(2x + y) + xy \cdot \frac{2}{2x + y} = y\log(2x + y) + \frac{2xy}{2x + y}$

* $y$ で偏微分：$\frac{\partial z}{\partial y} = x\log(2x + y) + xy \cdot \frac{1}{2x + y} = x\log(2x + y) + \frac{xy}{2x + y}$

3. 最終的な答え

(1)

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

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

(2)

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

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

(3)

$\frac{\partial z}{\partial x} = ye^{xy}\arctan(y)$

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

(4)

$\frac{\partial z}{\partial x} = y\log(2x + y) + \frac{2xy}{2x + y}$

$\frac{\partial z}{\partial y} = x\log(2x + y) + \frac{xy}{2x + y}$