a) $f(x, y) = x^2 y^3$ b) $f(x, y) = x^2 + y^2$ c) $f(x, y) = e^{x^2 + y^2}$ d) $f(x, y) = \ln(x^3 + y^3)$ e) $f(x, y) = (2x + 3y)(x^3 + y^3)$ f) $f(x, y) = \frac{4xy}{x^2 + y^2}$

はい、承知いたしました。画像の問題について、指示された形式で解答します。

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1. 問題の内容**

1. 以下の関数 $f(x, y)$ について、1階の偏導関数 $\frac{\partial f}{\partial x}$ と $\frac{\partial f}{\partial y}$ を求め、それぞれについて点 $(1, 2)$ における偏微分係数を求めよ。ただし、偏微分不可能な場合は「×」と答えよ。

a) $f(x, y) = x^2 y^3$

b) $f(x, y) = x^2 + y^2$

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

d) $f(x, y) = \ln(x^3 + y^3)$

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

f) $f(x, y) = \frac{4xy}{x^2 + y^2}$

2. 以下の関数 $f(x, y)$ について、2階の偏導関数 $\frac{\partial^2 f}{\partial x^2}$, $\frac{\partial^2 f}{\partial y^2}$, $\frac{\partial^2 f}{\partial x \partial y}$, $\frac{\partial^2 f}{\partial y \partial x}$ をすべて求めよ。

a) $f(x, y) = x^2 + 3xy + y^2$

b) $f(x, y) = \sqrt{x^2 + 2y^2}$

c) $f(x, y) = \ln(2x + y)$

d) $f(x, y) = 2xe^{x^2 + y^2}$

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2. 解き方の手順**

1. 偏導関数を求める：それぞれの関数 $f(x, y)$ に対して、変数 $x$ と $y$ について偏微分を行い、$\frac{\partial f}{\partial x}$ と $\frac{\partial f}{\partial y}$ を求めます。偏微分のルール（積の微分、合成関数の微分など）を適切に適用します。

2. 偏微分係数を求める：求めた偏導関数 $\frac{\partial f}{\partial x}$ と $\frac{\partial f}{\partial y}$ に、指定された点 $(1, 2)$ の $x$ 座標と $y$ 座標の値を代入します。これにより、点 $(1, 2)$ における偏微分係数が得られます。偏導関数が定義できない場合は「×」と答えます。

3. 2階の偏導関数を求める: それぞれの関数 $f(x,y)$ に対して、$\frac{\partial^2 f}{\partial x^2}$は$\frac{\partial f}{\partial x}$を$x$で偏微分、$\frac{\partial^2 f}{\partial y^2}$は$\frac{\partial f}{\partial y}$を$y$で偏微分、$\frac{\partial^2 f}{\partial x \partial y}$は$\frac{\partial f}{\partial y}$を$x$で偏微分、$\frac{\partial^2 f}{\partial y \partial x}$は$\frac{\partial f}{\partial x}$を$y$で偏微分することで求めます。

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3. 最終的な答え**

1. a) $f(x, y) = x^2 y^3$

* $\frac{\partial f}{\partial x} = 2xy^3$. $\frac{\partial f}{\partial x}(1, 2) = 2(1)(2^3) = 16$

* $\frac{\partial f}{\partial y} = 3x^2y^2$. $\frac{\partial f}{\partial y}(1, 2) = 3(1^2)(2^2) = 12$

b) $f(x, y) = x^2 + y^2$

* $\frac{\partial f}{\partial x} = 2x$. $\frac{\partial f}{\partial x}(1, 2) = 2(1) = 2$

* $\frac{\partial f}{\partial y} = 2y$. $\frac{\partial f}{\partial y}(1, 2) = 2(2) = 4$

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

* $\frac{\partial f}{\partial x} = 2xe^{x^2 + y^2}$. $\frac{\partial f}{\partial x}(1, 2) = 2(1)e^{1^2 + 2^2} = 2e^5$

* $\frac{\partial f}{\partial y} = 2ye^{x^2 + y^2}$. $\frac{\partial f}{\partial y}(1, 2) = 2(2)e^{1^2 + 2^2} = 4e^5$

d) $f(x, y) = \ln(x^3 + y^3)$

* $\frac{\partial f}{\partial x} = \frac{3x^2}{x^3 + y^3}$. $\frac{\partial f}{\partial x}(1, 2) = \frac{3(1^2)}{1^3 + 2^3} = \frac{3}{9} = \frac{1}{3}$

* $\frac{\partial f}{\partial y} = \frac{3y^2}{x^3 + y^3}$. $\frac{\partial f}{\partial y}(1, 2) = \frac{3(2^2)}{1^3 + 2^3} = \frac{12}{9} = \frac{4}{3}$

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

* $\frac{\partial f}{\partial x} = 2(x^3 + y^3) + (2x + 3y)(3x^2)$. $\frac{\partial f}{\partial x}(1, 2) = 2(1^3 + 2^3) + (2(1) + 3(2))(3(1^2)) = 2(9) + (8)(3) = 18 + 24 = 42$

* $\frac{\partial f}{\partial y} = 3(x^3 + y^3) + (2x + 3y)(3y^2)$. $\frac{\partial f}{\partial y}(1, 2) = 3(1^3 + 2^3) + (2(1) + 3(2))(3(2^2)) = 3(9) + (8)(12) = 27 + 96 = 123$

f) $f(x, y) = \frac{4xy}{x^2 + y^2}$

* $\frac{\partial f}{\partial x} = \frac{4y(x^2 + y^2) - 4xy(2x)}{(x^2 + y^2)^2} = \frac{4y^3 - 4x^2y}{(x^2 + y^2)^2}$. $\frac{\partial f}{\partial x}(1, 2) = \frac{4(2^3) - 4(1^2)(2)}{(1^2 + 2^2)^2} = \frac{32 - 8}{25} = \frac{24}{25}$

* $\frac{\partial f}{\partial y} = \frac{4x(x^2 + y^2) - 4xy(2y)}{(x^2 + y^2)^2} = \frac{4x^3 - 4xy^2}{(x^2 + y^2)^2}$. $\frac{\partial f}{\partial y}(1, 2) = \frac{4(1^3) - 4(1)(2^2)}{(1^2 + 2^2)^2} = \frac{4 - 16}{25} = -\frac{12}{25}$

2. a) $f(x, y) = x^2 + 3xy + y^2$

* $\frac{\partial f}{\partial x} = 2x + 3y$

* $\frac{\partial f}{\partial y} = 3x + 2y$

* $\frac{\partial^2 f}{\partial x^2} = 2$

* $\frac{\partial^2 f}{\partial y^2} = 2$

* $\frac{\partial^2 f}{\partial x \partial y} = 3$

* $\frac{\partial^2 f}{\partial y \partial x} = 3$

b) $f(x, y) = \sqrt{x^2 + 2y^2}$

* $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + 2y^2}}$

* $\frac{\partial f}{\partial y} = \frac{2y}{\sqrt{x^2 + 2y^2}}$

* $\frac{\partial^2 f}{\partial x^2} = \frac{\sqrt{x^2 + 2y^2} - x \frac{x}{\sqrt{x^2 + 2y^2}}}{x^2 + 2y^2} = \frac{2y^2}{(x^2 + 2y^2)^{3/2}}$

* $\frac{\partial^2 f}{\partial y^2} = \frac{2\sqrt{x^2 + 2y^2} - 2y \frac{2y}{\sqrt{x^2 + 2y^2}}}{x^2 + 2y^2} = \frac{2x^2}{(x^2 + 2y^2)^{3/2}}$

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

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

c) $f(x, y) = \ln(2x + y)$

* $\frac{\partial f}{\partial x} = \frac{2}{2x + y}$

* $\frac{\partial f}{\partial y} = \frac{1}{2x + y}$

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

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

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

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

d) $f(x, y) = 2xe^{x^2 + y^2}$

* $\frac{\partial f}{\partial x} = 2e^{x^2 + y^2} + 2x(2x)e^{x^2 + y^2} = (2 + 4x^2)e^{x^2 + y^2}$

* $\frac{\partial f}{\partial y} = 2x(2y)e^{x^2 + y^2} = 4xye^{x^2 + y^2}$

* $\frac{\partial^2 f}{\partial x^2} = 8xe^{x^2 + y^2} + (2 + 4x^2)(2x)e^{x^2 + y^2} = (12x + 8x^3)e^{x^2 + y^2}$

* $\frac{\partial^2 f}{\partial y^2} = 4x(x^2 + y^2)e^{x^2 + y^2} + 4xy(2y)e^{x^2 + y^2} = (4x^3 + 12xy^2)e^{x^2 + y^2}$

* $\frac{\partial^2 f}{\partial x \partial y} = 4ye^{x^2 + y^2} + 4xy(2x)e^{x^2 + y^2} = (4y + 8x^2y)e^{x^2 + y^2}$

* $\frac{\partial^2 f}{\partial y \partial x} = (4y + 8x^2y)e^{x^2 + y^2}$

これで問題の解答は以上となります。