第6章 6.11

(1)

z=x3−3axy+y3z = x^3 - 3axy + y^3

zx=3x2−3ayzy=−3ax+3y2z_x = 3x^2 - 3ay \qquad z_y = -3ax + 3y^2

zxx=6xzzy=−3azyy=6yz_{xx} = 6x \qquad z_{zy} = -3a \qquad z_{yy} = 6y

(2)

z=sin⁡xyz = \sin xy

zx=ycos⁡xyzy=xcos⁡xyz_x = y\cos xy \qquad z_y = x\cos xy

zxx=−y2sin⁡xyzxy=cos⁡xy−xysin⁡xyzyy=−x2sin⁡xyz_{xx} = -y^2 \sin xy \qquad z_{xy} = \cos xy - xy\sin xy \qquad z_{yy} = -x^2 \sin xy

(3)

z=x+yx−yz = \frac{x+y}{x-y}

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

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

(4)

z=sin⁡−1yx(x>0)z = \sin^{-1}\frac{y}{x} \qquad (x>0)

zx=−1x21y2−1x2=−yxx2−y2,zy=1x2−y2z_x = \frac{-\dfrac{1}{x^2}}{\sqrt{\dfrac{1}{y^2}-\dfrac{1}{x^2}}} = \frac{-y}{x\sqrt{x^2-y^2}}, \qquad z_y = \frac{1}{\sqrt{x^2-y^2}}

zxx=yx2x2−y2+yx⋅xx2−y2(x2−y2)2=(2x2−y2)yx2(x2−y2)32z_{xx} = \frac{\dfrac{y}{x^2}\sqrt{x^2-y^2} + \dfrac{y}{x}\cdot\dfrac{x}{\sqrt{x^2-y^2}}}{(\sqrt{x^2-y^2})^2} = \frac{(2x^2-y^2)y}{x^2(x^2-y^2)^{\frac{3}{2}}}

zxy=−x(x2−y2)3zyy=y(x2−y2)3z_{xy} = \frac{-x}{\sqrt{(x^2-y^2)^3}} \qquad z_{yy} = \frac{y}{\sqrt{(x^2-y^2)^3}}

解説: r31bn1z