6.2 偏導関数

1.

(a) $x$での偏微分は$x$以外の変数をすべて定数と見なして微分すればよいので，

$$f_{x}(x,y) = \frac{\partial (3x^{2})}{\partial x} - y\frac{\partial x}{\partial x} = 6x -y$$

$y$での偏微分は$y$以外の変数をすべて定数と見なして微分すればよいので，

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

(b) $x$での偏微分は$x$以外の変数をすべて定数と見なして微分すればよいので，

$$f_{x}(x,y) = e^{-y}\frac{\partial (x^{2})}{\partial x} = 2xe^{-y}$$

$y$での偏微分は$y$以外の変数をすべて定数と見なして微分すればよいので，

$$f_{y}(x,y) = x^{2}\frac{\partial (e^{-y})}{\partial y} = -x^{2}e^{-y}$$

(c) $x^{2}+y^{2}$を$t$とおくと

$$f_{x}(x,y) = \frac{\partial (\sqrt{t})}{\partial x} = \frac{\partial (\sqrt{t})}{\partial t}\frac{\partial t}{\partial x}$$

$$= \frac{1}{2\sqrt{t}}\cdot 2x = \frac{2x}{2\sqrt{x^{2}+y^{2}}} = \frac{x}{\sqrt{x^{2} + y^{2}}}$$

$$f_{y}(x,y) = \frac{\partial (\sqrt{t})}{\partial y} = \frac{\partial (\sqrt{t})}{\partial t}\frac{\partial t}{\partial y}$$

$$= \frac{1}{2\sqrt{t}}\cdot 2y = \frac{2y}{2\sqrt{x^{2}+y^{2}}} = \frac{y}{\sqrt{x^{2} + y^{2}}}$$

(d) $x$での偏微分は$x$以外の変数をすべて定数と見なして微分すればよいので，

$$f_{x}(x,y) = \sin{y}\frac{\partial x}{\partial x} = \sin{y}$$

$y$での偏微分は$y$以外の変数をすべて定数と見なして微分すればよいので，

$$f_{y}(x,y) = x\frac{\partial (\sin{y})}{\partial y} = x\sin{y}$$

(e) 商の微分法を用いると，

$$f_{x}(x,y) = \frac{\frac{\partial{(x-y)}}{\partial x}(x+y) - (x-y)\frac{\partial (x+y)}{\partial x}}{(x+y)^{2}}$$

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

$$f_{y}(x,y) = \frac{\frac{\partial{(x-y)}}{\partial y}(x+y) - (x-y)\frac{\partial (x+y)}{\partial y}}{(x+y)^{2}}$$

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

2.

(a)

$$f_{x} = \frac{\partial (ax^{2})}{\partial x} + 2by \frac{\partial x}{\partial x} = 2ax + 2by$$

$$f_{y} = 2bx\frac{\partial y}{\partial y} + \frac{\partial (cy^{2})}{\partial y} = 2bx + 2cy$$

$$f_{xx} = \frac{\partial (2ax)}{\partial x} = 2a$$

$$f_{xy} = \frac{\partial (2by)}{\partial y} = 2b$$

$$f_{yy} = \frac{\partial (2cy)}{\partial y} = 2c$$

(b)

$$f_{x} = 2(x + y^{2} + z^{3})\frac{\partial (x+y^{2}+z^{3})}{\partial x} = 2(x + y^{2} + z^{3})$$

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

$$f_{z} = 2(x + y^{2} + z^{3})\frac{\partial (x+y^{2}+z^{3})}{\partial z} = 6z^{2}(x + y^{2} + z^{3})$$

$$f_{xx} = \frac{\partial (2x)}{\partial x} = 2$$

$$f_{xy} = \frac{\partial (2y^{2})}{\partial y} = 4y$$

$$f_{xz} = \frac{\partial (2z^{3})}{\partial z} = 6z^{2}$$

$$f_{yy} = \frac{\partial (4y)}{\partial y}(x + y^{2} + z^{3}) + 4y\frac{\partial (y^{2})}{\partial y} = 4(x+ y^{2} + z^{3}) + 8y^{2}$$

$$f_{yz} = \frac{\partial (4yz^{3})}{\partial z} = 12yz^{2}$$

$$f_{zz} = \frac{\partial (6z^{2})}{\partial z}(x + y^{2} + z^{3}) + 6z^{2}\frac{\partial (z^{3})}{\partial z} = 12z(x + y^{2} + z^{3}) + 18z^{4}$$

(c)

$$f_{x} = \frac{\partial (\sin{t})}{\partial x} = \frac{\partial (\sin{t})}... ...artial (\sin{t})}{\partial t}\frac{\partial (3x-2y)}{\partial x} = 3\cos(3x-2y)$$

$$f_{y} = \frac{\partial (\sin{t})}{\partial y} = \frac{\partial (\sin{t})}... ...rtial (\sin{t})}{\partial t}\frac{\partial (3x-2y)}{\partial y} = -2\cos(3x-2y)$$

$$f_{xx} = \frac{\partial (3\cos(3x-2y))}{\partial x} = 3\frac{\partial (\cos(3x-2y))}{\partial x} = -9\sin(3x-2y)$$

$$f_{xy} = \frac{\partial (3\cos(3x-2y))}{\partial y} = 3\frac{\partial (\cos(3x-2y))}{\partial y} = 6\sin(3x-2y)$$

$$f_{yy} = -2\frac{\partial (\cos(3x-2y))}{\partial y} = -4\sin(3x-2y)$$

(d)

$$f_{x} = \frac{\partial (xe^{2y})}{\partial x} = e^{2y}\frac{\partial x}{\partial x} = e^{2y}$$

$$f_{y} = \frac{\partial (xe^{2y})}{\partial y} = x\frac{\partial (e^{2y})}{\partial y} = 2xe^{2y}$$

$$f_{xx} = \frac{\partial (e^{2y})}{\partial x} = 0$$

$$f_{xy} = \frac{\partial (e^{2y})}{\partial y} = 2e^{2y}$$

$$f_{yy} = 2x\frac{\partial (e^{2y})}{\partial y} = 4xe^{2y}$$