(1) $f(x,y) = \frac{x-y}{x+y}$ (2) $f(x,y) = e^{x^2y}$ (3) $f(x,y) = e^x \sin(2x^2+3y)$

## 数学の問題の解答

以下に、与えられた問題の解答を示します。

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1. 問題の内容

以下の問題が与えられています。

1. 関数 $f(x,y)$ を偏微分し、$x$ および $y$ についての偏導関数 $f_x$ および $f_y$ を求める。

(1) $f(x,y) = \frac{x-y}{x+y}$

(2) $f(x,y) = e^{x^2y}$

(3) $f(x,y) = e^x \sin(2x^2+3y)$

2. $z = x^2 + y^3$, $x = 3t^2 + 2t + 1$, $y = -2t - 3$ のとき、$\frac{dz}{dt}$ を求める。

3. $z = x^2 - 2xy + y^2$, $x = r \cos \theta$, $y = r \sin \theta$ のとき、$\frac{\partial z}{\partial r}$, $\frac{\partial z}{\partial \theta}$ を求める。

4. $f(x,y) = \log \sqrt{x^2 + y^2}$ について、$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ を求める。

5. $z = f(x,y)$, $x = r \cos \theta$, $y = r \sin \theta$ のとき、次の等式が成り立つことを証明する。

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

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2. 解き方の手順

**

1. (1) $f(x,y) = \frac{x-y}{x+y}$**

* $f_x$ を求める:

商の微分公式を用いる。 $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$

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

* $f_y$ を求める:

同様に商の微分公式を用いる。

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

**(2) $f(x,y) = e^{x^2y}$**

* $f_x$ を求める:

合成関数の微分を用いる。

$f_x = e^{x^2y} \cdot (2xy) = 2xy e^{x^2y}$

* $f_y$ を求める:

合成関数の微分を用いる。

$f_y = e^{x^2y} \cdot (x^2) = x^2 e^{x^2y}$

**(3) $f(x,y) = e^x \sin(2x^2+3y)$**

* $f_x$ を求める:

積の微分と合成関数の微分を用いる。

$f_x = e^x \sin(2x^2+3y) + e^x \cos(2x^2+3y) \cdot (4x) = e^x \sin(2x^2+3y) + 4xe^x \cos(2x^2+3y)$

* $f_y$ を求める:

合成関数の微分を用いる。

$f_y = e^x \cos(2x^2+3y) \cdot (3) = 3e^x \cos(2x^2+3y)$

**

2. $z = x^2 + y^3$, $x = 3t^2 + 2t + 1$, $y = -2t - 3$**

* $\frac{dz}{dt}$ を求める:

合成関数の微分を用いる。 $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$

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

$\frac{dx}{dt} = 6t + 2$, $\frac{dy}{dt} = -2$

したがって、

$\frac{dz}{dt} = 2x(6t+2) + 3y^2(-2) = 2(3t^2+2t+1)(6t+2) - 6(-2t-3)^2$

$= 2(18t^3 + 12t^2 + 6t + 6t^2 + 4t + 2) - 6(4t^2 + 12t + 9) = 36t^3 + 36t^2 + 20t + 4 - 24t^2 - 72t - 54$

$= 36t^3 + 12t^2 - 52t - 50$

**

3. $z = x^2 - 2xy + y^2$, $x = r \cos \theta$, $y = r \sin \theta$**

* $\frac{\partial z}{\partial r}$ を求める:

合成関数の微分を用いる。 $\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$

$\frac{\partial z}{\partial x} = 2x - 2y$, $\frac{\partial z}{\partial y} = -2x + 2y$

$\frac{\partial x}{\partial r} = \cos \theta$, $\frac{\partial y}{\partial r} = \sin \theta$

したがって、

$\frac{\partial z}{\partial r} = (2x - 2y)\cos \theta + (-2x + 2y)\sin \theta = 2(x-y)\cos \theta - 2(x-y)\sin \theta = 2(x-y)(\cos \theta - \sin \theta)$

ここで、$x = r \cos \theta$, $y = r \sin \theta$ を代入すると、

$\frac{\partial z}{\partial r} = 2(r \cos \theta - r \sin \theta)(\cos \theta - \sin \theta) = 2r(\cos \theta - \sin \theta)^2 = 2r(\cos^2 \theta - 2\sin \theta \cos \theta + \sin^2 \theta) = 2r(1 - 2\sin \theta \cos \theta) = 2r(1 - \sin 2\theta)$

* $\frac{\partial z}{\partial \theta}$ を求める:

合成関数の微分を用いる。 $\frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}$

$\frac{\partial z}{\partial x} = 2x - 2y$, $\frac{\partial z}{\partial y} = -2x + 2y$

$\frac{\partial x}{\partial \theta} = -r \sin \theta$, $\frac{\partial y}{\partial \theta} = r \cos \theta$

したがって、

$\frac{\partial z}{\partial \theta} = (2x - 2y)(-r \sin \theta) + (-2x + 2y)(r \cos \theta) = -2r(x-y)\sin \theta - 2r(x-y)\cos \theta = -2r(x-y)(\sin \theta + \cos \theta)$

ここで、$x = r \cos \theta$, $y = r \sin \theta$ を代入すると、

$\frac{\partial z}{\partial \theta} = -2r(r \cos \theta - r \sin \theta)(\sin \theta + \cos \theta) = -2r^2(\cos \theta - \sin \theta)(\sin \theta + \cos \theta) = -2r^2(\cos^2 \theta - \sin^2 \theta) = -2r^2 \cos 2\theta$

**

4. $f(x,y) = \log \sqrt{x^2 + y^2}$**

* $\frac{\partial f}{\partial x}$ を求める:

$f(x,y) = \log (x^2+y^2)^{1/2} = \frac{1}{2} \log (x^2+y^2)$

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

* $\frac{\partial^2 f}{\partial x^2}$ を求める:

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

* $\frac{\partial f}{\partial y}$ を求める:

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

* $\frac{\partial^2 f}{\partial y^2}$ を求める:

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

* $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ を求める:

$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{y^2 - x^2}{(x^2+y^2)^2} + \frac{x^2 - y^2}{(x^2+y^2)^2} = 0$

**

5. $z = f(x,y)$, $x = r \cos \theta$, $y = r \sin \theta$**

* $\frac{\partial z}{\partial r}$ を求める:

$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r} = \frac{\partial z}{\partial x} \cos \theta + \frac{\partial z}{\partial y} \sin \theta$

* $\frac{\partial z}{\partial \theta}$ を求める:

$\frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta} = \frac{\partial z}{\partial x}(-r \sin \theta) + \frac{\partial z}{\partial y}(r \cos \theta)$

* $(\frac{\partial z}{\partial r})^2$ を求める:

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

* $(\frac{\partial z}{\partial \theta})^2$ を求める:

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

* $(\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2$ を求める:

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

したがって、 $(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2$

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3. 最終的な答え

1. (1) $f_x = \frac{2y}{(x+y)^2}$, $f_y = \frac{-2x}{(x+y)^2}$

(2) $f_x = 2xye^{x^2y}$, $f_y = x^2e^{x^2y}$

(3) $f_x = e^x \sin(2x^2+3y) + 4xe^x \cos(2x^2+3y)$, $f_y = 3e^x \cos(2x^2+3y)$

2. $\frac{dz}{dt} = 36t^3 + 12t^2 - 52t - 50$

3. $\frac{\partial z}{\partial r} = 2r(1 - \sin 2\theta)$, $\frac{\partial z}{\partial \theta} = -2r^2 \cos 2\theta$

4. $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$

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