与えられた4つの関数 $f(x, y)$ のそれぞれについて、1階偏導関数 $f_x$, $f_y$ と、2階偏導関数 $f_{xx}$, $f_{yy}$, $f_{xy}$, $f_{yx}$ を求める。

はい、承知いたしました。画像に記載された4つの関数について、それぞれ2階までの偏導関数を求めます。

1. 問題の内容

与えられた4つの関数 $f(x, y)$ のそれぞれについて、1階偏導関数 $f_x$, $f_y$ と、2階偏導関数 $f_{xx}$, $f_{yy}$, $f_{xy}$, $f_{yx}$ を求める。

2. 解き方の手順

各関数について、以下の手順で偏導関数を求めます。

(1) $f(x, y) = x^4 + xy^3$

* $f_x = \frac{\partial f}{\partial x} = 4x^3 + y^3$

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

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

* $f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (3xy^2) = 6xy$

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

* $f_{yx} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (4x^3 + y^3) = 3y^2$

(2) $f(x, y) = e^x \cos(y)$

* $f_x = \frac{\partial f}{\partial x} = e^x \cos(y)$

* $f_y = \frac{\partial f}{\partial y} = -e^x \sin(y)$

* $f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (e^x \cos(y)) = e^x \cos(y)$

* $f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (-e^x \sin(y)) = -e^x \cos(y)$

* $f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (-e^x \sin(y)) = -e^x \sin(y)$

* $f_{yx} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (e^x \cos(y)) = -e^x \sin(y)$

(3) $f(x, y) = \frac{\sqrt{x+y}}{xy}$

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

* $f_x = \frac{\frac{1}{2}(x+y)^{-1/2}xy - (x+y)^{1/2}y}{x^2y^2} = \frac{\frac{xy}{2\sqrt{x+y}} - y\sqrt{x+y}}{x^2y^2} = \frac{x - 2(x+y)}{2x^2y\sqrt{x+y}} = \frac{-x-2y}{2x^2y\sqrt{x+y}}$

* $f_y = \frac{\frac{1}{2}(x+y)^{-1/2}xy - (x+y)^{1/2}x}{x^2y^2} = \frac{\frac{xy}{2\sqrt{x+y}} - x\sqrt{x+y}}{x^2y^2} = \frac{y - 2(x+y)}{2xy^2\sqrt{x+y}} = \frac{-2x-y}{2xy^2\sqrt{x+y}}$

* $f_{xx} = \frac{\partial f_x}{\partial x} = \frac{\partial}{\partial x} \left(\frac{-x-2y}{2x^2y\sqrt{x+y}}\right)$

* $f_{yy} = \frac{\partial f_y}{\partial y} = \frac{\partial}{\partial y} \left(\frac{-2x-y}{2xy^2\sqrt{x+y}}\right)$

* $f_{xy} = \frac{\partial f_x}{\partial y} = \frac{\partial}{\partial y} \left(\frac{-x-2y}{2x^2y\sqrt{x+y}}\right)$

* $f_{yx} = \frac{\partial f_y}{\partial x} = \frac{\partial}{\partial x} \left(\frac{-2x-y}{2xy^2\sqrt{x+y}}\right)$

(4) $f(x, y) = e^x \sin(y)$

* $f_x = \frac{\partial f}{\partial x} = e^x \sin(y)$

* $f_y = \frac{\partial f}{\partial y} = e^x \cos(y)$

* $f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (e^x \sin(y)) = e^x \sin(y)$

* $f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (e^x \cos(y)) = -e^x \sin(y)$

* $f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (e^x \cos(y)) = e^x \cos(y)$

* $f_{yx} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (e^x \sin(y)) = e^x \cos(y)$

3. 最終的な答え

(1)

$f_x = 4x^3 + y^3$

$f_y = 3xy^2$

$f_{xx} = 12x^2$

$f_{yy} = 6xy$

$f_{xy} = 3y^2$

$f_{yx} = 3y^2$

(2)

$f_x = e^x \cos(y)$

$f_y = -e^x \sin(y)$

$f_{xx} = e^x \cos(y)$

$f_{yy} = -e^x \cos(y)$

$f_{xy} = -e^x \sin(y)$

$f_{yx} = -e^x \sin(y)$

(3)

$f_x = \frac{-x-2y}{2x^2y\sqrt{x+y}}$

$f_y = \frac{-2x-y}{2xy^2\sqrt{x+y}}$

$f_{xx}$, $f_{yy}$, $f_{xy}$, $f_{yx}$ は複雑なため省略

(4)

$f_x = e^x \sin(y)$

$f_y = e^x \cos(y)$

$f_{xx} = e^x \sin(y)$

$f_{yy} = -e^x \sin(y)$

$f_{xy} = e^x \cos(y)$

$f_{yx} = e^x \cos(y)$