6.3 解答

$\displaystyle z_{x}$	$\displaystyle =$	$\displaystyle \frac{\partial (x)}{\partial x}y^2 e^{\frac{x}{y}} + xy^2 \frac{\partial (e^{x/y})}{\partial x}$
	$\displaystyle =$	$\displaystyle y^2 e^{\frac{x}{y}} + xy^2 \frac{1}{y}e^{\frac{x}{y}} = (y^2 + xy)e^{\frac{x}{y}}$
$\displaystyle z_{y}$	$\displaystyle =$	$\displaystyle x \frac{\partial (y)}{\partial y} e^{\frac{x}{y}} + xy^2 \frac{\partial (e^{x/y})}{\partial y}$
	$\displaystyle =$	$\displaystyle 2xy e^{\frac{x}{y}} + xy^2 e^{\frac{x}{y}}(-\frac{x}{y^2}) = (2xy - x^2)e^{\frac{x}{y}}$
$\displaystyle z_{xx}$	$\displaystyle =$	$\displaystyle \frac{\partial (y^2 + xy)}{\partial x}e^{\frac{x}{y}} + (y^2 + xy)\frac{\partial (e^{x/y})}{\partial x}$
	$\displaystyle =$	$\displaystyle ye^{\frac{x}{y}} + (y^2 + xy)e^{\frac{x}{y}}(\frac{\partial(\frac{x}{y})}{\partial x})$
	$\displaystyle =$	$\displaystyle ye^{\frac{x}{y}} + (y^2 + xy)(\frac{1}{y})e^{\frac{x}{y}}$
	$\displaystyle =$	$\displaystyle e^{\frac{x}{y}}(y + y + x) = e^{\frac{x}{y}}(x + 2y)$
$\displaystyle z_{xy}$	$\displaystyle =$	$\displaystyle \frac{\partial (y^2 + xy)}{\partial y}e^{\frac{x}{y}} + (y^2 + xy)\frac{\partial (e^{x/y})}{\partial y}$
	$\displaystyle =$	$\displaystyle (2y+x)e^{\frac{x}{y}} + (y^2 + xy)e^{\frac{x}{y}}(\frac{\partial(\frac{x}{y})}{\partial y})$
	$\displaystyle =$	$\displaystyle (2y + x)e^{\frac{x}{y}} + (y^2 + xy)(-\frac{x}{y^2})e^{\frac{x}{y}}$
	$\displaystyle =$	$\displaystyle e^{\frac{x}{y}}(2y + x - x - \frac{x}{y^2}) = e^{\frac{x}{y}}(2y - \frac{x^2}{y})$
$\displaystyle z_{yy}$	$\displaystyle =$	$\displaystyle \frac{\partial (2xy - x^2)}{\partial y}e^{\frac{x}{y}} + (2xy - x^2)\frac{\partial (e^{x/y})}{\partial y}$
	$\displaystyle =$	$\displaystyle 2xe^{\frac{x}{y}} + (2xy - x^2)e^{\frac{x}{y}}(\frac{\partial(\frac{x}{y})}{\partial y})$
	$\displaystyle =$	$\displaystyle 2xe^{\frac{x}{y}} + (2xy - x^2)(-\frac{x}{y^2})e^{\frac{x}{y}}$
	$\displaystyle =$	$\displaystyle e^{\frac{x}{y}}(2x - \frac{2x^2}{y} + \frac{x^3}{y^2})$

$\displaystyle z_{x}$	$\displaystyle =$	$\displaystyle \frac{1}{1 + (x^2 + y^2)^2}\frac{\partial(x^2 + y^2)}{\partial x}$
	$\displaystyle =$	$\displaystyle \frac{2x}{1 + (x^2 + y^2)^2}$
$\displaystyle z_{y}$	$\displaystyle =$	$\displaystyle \frac{1}{1 + (x^2 + y^2)^2}\frac{\partial(x^2 + y^2)}{\partial y}$
	$\displaystyle =$	$\displaystyle \frac{2y}{1 + (x^2 + y^2)^2}$
$\displaystyle z_{xx}$	$\displaystyle =$	$\displaystyle \frac{\frac{\partial(2x)}{\partial x}(1 + (x^2 + y^2)^2) - 2x(\frac{\partial(1 + (x^2 + y^2)^2)}{\partial x})}{(1 + (x^2 + y^2)^2)^2}$
	$\displaystyle =$	$\displaystyle \frac{2(1 + (x^2 + y^2)^2) - 2x(2(x^2 + y^2)(2x))}{(1 + (x^2 +y^2)^2)^2}$
	$\displaystyle =$	$\displaystyle \frac{2 + 2(x^2 + y^2)^2 - 8x^2(x^2 + y^2)}{(1 + (x^2 + y^2)^2)^2}$
$\displaystyle z_{xy}$	$\displaystyle =$	$\displaystyle -\frac{2x(2(x^2 + y^2)(2x))}{(1 + (x^2 +y^2)^2)^2}$
	$\displaystyle =$	$\displaystyle \frac{ - 8x(x^2 + y^2)}{(1 + (x^2 + y^2)^2)^2}$
$\displaystyle z_{yy}$	$\displaystyle =$	$\displaystyle \frac{\frac{\partial(2y)}{\partial y}(1 + (x^2 + y^2)^2) - 2y(\frac{\partial(1 + (x^2 + y^2)^2)}{\partial y})}{(1 + (x^2 + y^2)^2)^2}$
	$\displaystyle =$	$\displaystyle \frac{2 + 2(x^2 + y^2)^2 - 8y^2(x^2 + y^2)}{(1 + (x^2 + y^2)^2)}$

(a) $f_{x}(0,0), f_{y}(0,0)$ が存在するか調べればよい． $f{x}(x,0) = \frac{0}{x^2} = 0$ より $f_{x}(x,0) = 0$ ．したがって， $f_{x}(0,0) = 0$ ． $f{y}(0,y) = \frac{y^3}{y^2} = y$ ． $f_{y}(0,y) = 1$ ．したがって， $f_{y}(0,0) = 1$ ．

(b) $f(x,0) = \log{1} = 0$ より $f_{x}(x,0) = 0$ . したがって， $f_{x}(0,0) = 0$ . $f(0,y) = \log(1+y^2)$ より $f_{y}(0,y) = \frac{2y}{1 + y^2}$ ．したがって， $f_{y}(0,0) = 0$ .