与えられた12個の関数 $z(x, y)$ を、それぞれ $x$ と $y$ について偏微分する問題です。

1. 問題の内容

与えられた12個の関数 $z(x, y)$ を、それぞれ $x$ と $y$ について偏微分する問題です。

2. 解き方の手順

(1) $z = 4x + 5y$

* $x$ で偏微分: $\frac{\partial z}{\partial x} = 4$

* $y$ で偏微分: $\frac{\partial z}{\partial y} = 5$

(2) $z = x^3 - 3y^2$

* $x$ で偏微分: $\frac{\partial z}{\partial x} = 3x^2$

* $y$ で偏微分: $\frac{\partial z}{\partial y} = -6y$

(3) $z = x^2 y^3$

* $x$ で偏微分: $\frac{\partial z}{\partial x} = 2xy^3$

* $y$ で偏微分: $\frac{\partial z}{\partial y} = 3x^2y^2$

(4) $z = x^2y - 5xy$

* $x$ で偏微分: $\frac{\partial z}{\partial x} = 2xy - 5y$

* $y$ で偏微分: $\frac{\partial z}{\partial y} = x^2 - 5x$

(5) $z = e^x \sin y$

* $x$ で偏微分: $\frac{\partial z}{\partial x} = e^x \sin y$

* $y$ で偏微分: $\frac{\partial z}{\partial y} = e^x \cos y$

(6) $z = (x^2 + y)(x - 3y)$

* $x$ で偏微分: $\frac{\partial z}{\partial x} = 2x(x - 3y) + (x^2 + y)(1) = 2x^2 - 6xy + x^2 + y = 3x^2 - 6xy + y$

* $y$ で偏微分: $\frac{\partial z}{\partial y} = (1)(x - 3y) + (x^2 + y)(-3) = x - 3y - 3x^2 - 3y = x - 3x^2 - 6y$

(7) $z = (x^2 - y) \cos x$

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

* $y$ で偏微分: $\frac{\partial z}{\partial y} = (-1) \cos x = -\cos x$

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

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

* $y$ で偏微分: $\frac{\partial z}{\partial y} = \frac{0(x^2-y)-x(-1)}{(x^2 - y)^2} = \frac{x}{(x^2 - y)^2}$

(9) $z = \frac{\sin y}{\cos x}$

* $x$ で偏微分: $\frac{\partial z}{\partial x} = \frac{\sin y \cdot \sin x}{\cos^2 x} = \frac{\sin y}{\cos x} \cdot \frac{\sin x}{\cos x} = \tan x \frac{\sin y}{\cos x}$

* $y$ で偏微分: $\frac{\partial z}{\partial y} = \frac{\cos y}{\cos x}$

(10) $z = (3x^2 + y)^4$

* $x$ で偏微分: $\frac{\partial z}{\partial x} = 4(3x^2 + y)^3 (6x) = 24x (3x^2 + y)^3$

* $y$ で偏微分: $\frac{\partial z}{\partial y} = 4(3x^2 + y)^3 (1) = 4(3x^2 + y)^3$

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

* $x$ で偏微分: $\frac{\partial z}{\partial x} = \frac{4x^3}{x^4 + y^2}$

* $y$ で偏微分: $\frac{\partial z}{\partial y} = \frac{2y}{x^4 + y^2}$

(12) $z = \sqrt{x^2 + y^2} = (x^2 + y^2)^{1/2}$

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

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

3. 最終的な答え

(1) $\frac{\partial z}{\partial x} = 4$, $\frac{\partial z}{\partial y} = 5$

(2) $\frac{\partial z}{\partial x} = 3x^2$, $\frac{\partial z}{\partial y} = -6y$

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

(4) $\frac{\partial z}{\partial x} = 2xy - 5y$, $\frac{\partial z}{\partial y} = x^2 - 5x$

(5) $\frac{\partial z}{\partial x} = e^x \sin y$, $\frac{\partial z}{\partial y} = e^x \cos y$

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

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

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

(9) $\frac{\partial z}{\partial x} = \tan x \frac{\sin y}{\cos x}$, $\frac{\partial z}{\partial y} = \frac{\cos y}{\cos x}$

(10) $\frac{\partial z}{\partial x} = 24x (3x^2 + y)^3$, $\frac{\partial z}{\partial y} = 4(3x^2 + y)^3$

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

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