6.3 Answer

6.3

1.

(a)

$$z_{x} = 3x^2 + y^2, z_{y} = 2xy + 3y^2$$

(b)

$$z_{x} = \frac{\partial(e^{x})}{\partial x}\sin{y} = e^{x}\sin{y},$$

$$z_{y} = e^{x}\frac{\partial(\sin{y})}{\partial y} = e^{x}\cos{y}.$$

(c)

$$z_{x} = \frac{1}{x^2 + y^2}\frac{\partial(x^2 + y^2)}{\partial x} = \frac{2x}{x^2 + y^2}$$

$$z_{y} = \frac{1}{x^2 + y^2}\frac{\partial(x^2 + y^2)}{\partial y} = \frac{2y}{x^2 + y^2}$$

2.

(a)

$$z_{x} = \frac{\partial(x^3)}{\partial x}y + \frac{\partial(x)}{\partial x}y^2 = 3x^2 y + y^2,$$

$$z_{y} = x^3 \frac{\partial(y)}{\partial y} + x \frac{\partial(y^2)}{\partial y} = x^3 + 2xy$$

$$z_{xx} = 6xy, z_{xy} = 3x^2 + 2y, z_{yx} = 3x^2 + 2y, z_{yy} = 2x$$

(b)

$$z_{x} = \frac{\partial (x)}{\partial x}y^2 e^{\frac{x}{y}} + xy^2 \frac{\partial (e^{x/y})}{\partial x}$$

$$= y^2 e^{\frac{x}{y}} + xy^2 \frac{1}{y}e^{\frac{x}{y}} = (y^2 + xy)e^{\frac{x}{y}}$$

$$z_{y} = x \frac{\partial (y)}{\partial y} e^{\frac{x}{y}} + xy^2 \frac{\partial (e^{x/y})}{\partial y}$$

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

$$z_{xx} = \frac{\partial (y^2 + xy)}{\partial x}e^{\frac{x}{y}} + (y^2 + xy)\frac{\partial (e^{x/y})}{\partial x}$$

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

$$= ye^{\frac{x}{y}} + (y^2 + xy)(\frac{1}{y})e^{\frac{x}{y}}$$

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

$$z_{xy} = \frac{\partial (y^2 + xy)}{\partial y}e^{\frac{x}{y}} + (y^2 + xy)\frac{\partial (e^{x/y})}{\partial y}$$

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

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

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

$$z_{yy} = \frac{\partial (2xy - x^2)}{\partial y}e^{\frac{x}{y}} + (2xy - x^2)\frac{\partial (e^{x/y})}{\partial y}$$

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

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

$$= e^{\frac{x}{y}}(2x - \frac{2x^2}{y} + \frac{x^3}{y^2})$$

(c)

$$z_{x} = \frac{1}{1 + (x^2 + y^2)^2}\frac{\partial(x^2 + y^2)}{\partial x}$$

$$= \frac{2x}{1 + (x^2 + y^2)^2}$$

$$z_{y} = \frac{1}{1 + (x^2 + y^2)^2}\frac{\partial(x^2 + y^2)}{\partial y}$$

$$= \frac{2y}{1 + (x^2 + y^2)^2}$$

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

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

$$= \frac{2 + 2(x^2 + y^2)^2 - 8x^2(x^2 + y^2)}{(1 + (x^2 + y^2)^2)^2}$$

$$z_{xy} = -\frac{2x(2(x^2 + y^2)(2x))}{(1 + (x^2 +y^2)^2)^2}$$

$$= \frac{ - 8x(x^2 + y^2)}{(1 + (x^2 + y^2)^2)^2}$$

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

$$= \frac{2 + 2(x^2 + y^2)^2 - 8y^2(x^2 + y^2)}{(1 + (x^2 + y^2)^2)}$$

3.

(a) We need to find $f_{x}(0,0), f_{y}(0,0)$ exist. $f{x}(x,0) = \frac{0}{x^2} = 0$より $f_{x}(x,0) = 0$．Thus， $f_{x}(0,0) = 0$． $f{y}(0,y) = \frac{y^3}{y^2} = y$． $f_{y}(0,y) = 1$．Therefore， $f_{y}(0,0) = 1$．

(b) $f(x,0) = \log{1} = 0$. Thus, $f_{x}(x,0) = 0$. Then， $f_{x}(0,0) = 0$. $f(0,y) = \log(1+y^2)$より $f_{y}(0,y) = \frac{2y}{1 + y^2}$．Therefore,， $f_{y}(0,0) = 0$.