第6章 6.4

(1)

z=x3+3x2y+y3z = x^3 + 3x^2 y + y^3

zx=3x2+6xyzy=3x2+3y2z_x = 3x^2 + 6xy \qquad z_y = 3x^2 + 3y^2

zxx=6x+6yzxy=6xzyy=6yz_{xx} = 6x + 6y \qquad z_{xy} = 6x \qquad z_{yy} = 6y

(2)

z=(x2+y)3z = (x^2+y)^3

zx=3(x2+y)2⋅2x=6x(x2+y)2zy=3(x2+y)2z_x = 3(x^2+y)^2 \cdot 2x = 6x(x^2+y)^2 \qquad z_y = 3(x^2+y)^2

zxx=6(x2+y)2+6x⋅2(x2+y)⋅2x=(x2+y)(6x2+6y+24x2)=(x2+y)(30x2+6y)z_{xx} = 6(x^2+y)^2 + 6x \cdot 2(x^2+y) \cdot 2x = (x^2+y)(6x^2+6y+24x^2) = (x^2+y)(30x^2+6y)

zxy=12x(x2+y)zyy=6(x2+y)z_{xy} = 12x(x^2+y) \qquad z_{yy} = 6(x^2+y)

(3)

z=xx+y=1−yx+yz = \frac{x}{x+y} = 1 - \frac{y}{x+y}

zx=y(x+y)2zy=−x(x+y)2z_x = \frac{y}{(x+y)^2} \qquad z_y = -\frac{x}{(x+y)^2}

zxx=−2y(x+y)3zxy=(x+y)2−y⋅2(x+y)(x+y)4=(x+y)(x+y−2y)(x+y)4=x−y(x+y)3z_{xx} = \frac{-2y}{(x+y)^3} \qquad z_{xy} = \frac{(x+y)^2 - y \cdot 2(x+y)}{(x+y)^4} = \frac{(x+y)(x+y-2y)}{(x+y)^4} = \frac{x-y}{(x+y)^3}

zyy=2x(x+y)3z_{yy} = \frac{2x}{(x+y)^3}

(4)

z=log⁡x2yz = \log x^2 y

zx=2xyx2y=2xzy=x2x2y=1yz_x = \frac{2xy}{x^2 y} = \frac{2}{x} \qquad z_y = \frac{x^2}{x^2 y} = \frac{1}{y}

zxx=−2x2zxy=0zyy=−1y2z_{xx} = -\frac{2}{x^2} \qquad z_{xy} = 0 \qquad z_{yy} = -\frac{1}{y^2}

(5)

z=e2xcos⁡3yz = e^{2x} \cos 3y

zx=2e2xcos⁡3yzy=−3e2xsin⁡3yz_x = 2e^{2x} \cos 3y \qquad z_y = -3e^{2x} \sin 3y

zxx=4e2xcos⁡3yzxy=−6e2xsin⁡3yzyy=−9e2xcos⁡3yz_{xx} = 4e^{2x} \cos 3y \qquad z_{xy} = -6e^{2x} \sin 3y \qquad z_{yy} = -9e^{2x} \cos 3y

(6)

z=sin⁡x⋅cos⁡yz = \sin x \cdot \cos y

zx=cos⁡x⋅cos⁡yzy=−sin⁡x⋅sin⁡yz_x = \cos x \cdot \cos y \qquad z_y = -\sin x \cdot \sin y

zxx=−sin⁡x⋅cos⁡yzxy=−cos⁡x⋅sin⁡yzyy=−sin⁡x⋅cos⁡yz_{xx} = -\sin x \cdot \cos y \qquad z_{xy} = -\cos x \cdot \sin y \qquad z_{yy} = -\sin x \cdot \cos y

(7)

z=xy(x>0)z = x^y \qquad (x > 0)

zx=yxy−1zy=xylog⁡xz_x = yx^{y-1} \qquad z_y = x^y \log x

zxx=y(y−1)xy−2zxy=xy−1+yxy−1log⁡xzyy=xy(log⁡x)2z_{xx} = y(y-1)x^{y-2} \qquad z_{xy} = x^{y-1} + yx^{y-1}\log x \qquad z_{yy} = x^y(\log x)^2

(8)

z=tan⁡−1y2xz = \tan^{-1}\frac{y}{2x}

zx=−2y4x21+(y2x)2=−2y4x2+y2zy=2xy2+4x2z_x = \frac{-\dfrac{2y}{4x^2}}{1+\left(\dfrac{y}{2x}\right)^2} = \frac{-2y}{4x^2+y^2} \qquad z_y = \frac{2x}{y^2+4x^2}

zxx=16xy(4x2+y2)2zxy=−2(4x2+y2)+2y⋅2y(4x2+y2)2=−8x2+2y2(4x2+y2)2zyy=−4xy(y2+4x2)2z_{xx} = \frac{16xy}{(4x^2+y^2)^2} \qquad z_{xy} = \frac{-2(4x^2+y^2)+2y \cdot 2y}{(4x^2+y^2)^2} = \frac{-8x^2+2y^2}{(4x^2+y^2)^2} \qquad z_{yy} = \frac{-4xy}{(y^2+4x^2)^2}

解説: r31bn1z