$3$\displaystyle{z=x- 2y}$	解答	$3$\displaystyle{z={x}^{2}+{y}^{2}}$	解答
$3$\displaystyle{z={x}^{2}- 3xy+2{y}^{2}}$	解答	$3$\displaystyle{z=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$	解答
$3$\displaystyle{z=\frac{\hspace{2} x- y \hspace{2}}{\hspace{2} x+y \hspace{2}}}$	解答	$3$\displaystyle{z=\sqrt{ 3x- 4y }}$	解答
$3$\displaystyle{z={e}^{ xy }}$	解答	$3$\displaystyle{z=\log $ x- y $ }$	解答
$3$\displaystyle{z=2\sin \sqrt{ xy }}$	解答	$3$\displaystyle{z={\tan }^{ - 1 }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$	解答
$3$\displaystyle{z={x}^{3}+3{x}^{2}y+2{y}^{3}}$	解答	$3$\displaystyle{z={e}^{ - ax }\text{\hspace{1}\hspace{1}} $ \sin by+\cos by $ }$	解答
$3$\displaystyle{f $ x,y $ =\frac{\hspace{2} 5x+3y \hspace{2}}{\hspace{2} 3x+2y \hspace{2}}}$	解答	$3$\displaystyle{f $ x,y $ ={x}^{y}}$	解答

次の関数について $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}f $ 1,- 2 $ \text{\hspace{1}}\text{\hspace{1}}}$ と $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}f $ - 1,2 $ \text{\hspace{1}}\text{\hspace{1}}}$ を求めよ．

	$3$\displaystyle{f $ x,y $ ={x}^{2}+xy- {y}^{2}}$	解答		$3$\displaystyle{f $ x,y $ =\sqrt{ {x}^{2}- xy }}$	解答
	$3$\displaystyle{ f $ x,y $ ={ \sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} }$	解答

次のことを証明せよ．

	$3$\displaystyle{z=f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ \text{\hspace{1}}\text{\hspace{1}}}$ ならば $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=0\text{\hspace{1}}\text{\hspace{1}}}$ である．	解答
	$3$\displaystyle{z=f $ {x}^{2}- {y}^{2} $ \text{\hspace{1}\hspace{1}}}$ ならば $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=0\text{\hspace{1} }\text{\hspace{1} }}$ である．	解答
	$3$\displaystyle{z=\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ \text{\hspace{1} }\text{\hspace{1} }}$ ならば $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}+z=0\text{\hspace{1} }\text{\hspace{1} }}$ である．	解答
	$3$\displaystyle{z=f $ {x}^{2}- {y}^{2} $ }$ ならば $3$\displaystyle{y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=0}$ である．	解答
	$3$\displaystyle{z=\log \sqrt{ {x}^{2}+{y}^{2} }\text{\hspace{1} }\text{\hspace{1} }}$ ならば $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }{ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ }^{2}+{ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} {e}^{ 2z } \hspace{2}}\text{\hspace{1} }\text{\hspace{1} }}$ である．	解答

次の関数の $3$\displaystyle{\frac{\hspace{2} dz \hspace{2}}{\hspace{2} dt \hspace{2}}}$ を求めよ．

	$3$\displaystyle{z=f $ x,y $ \text{\hspace{1} },}$ $3$\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=a+ht\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=b+kt\text{\hspace{1} }\text{\hspace{1} }}$	解答		$3$\displaystyle{z={x}^{2}+{y}^{2}\text{\hspace{1} },}$ $3$\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=t- \sin t\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=1- \cos t\text{\hspace{1} }\text{\hspace{1} }}$	解答
	$3$\displaystyle{z=xy\text{\hspace{1} },}$ $3$\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=2{t}^{2}+1\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y={t}^{2}+3t+1\text{\hspace{1} }\text{\hspace{1} }}$	解答		$3$\displaystyle{z=x\tan y\text{\hspace{1} },}$ $3$\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x={\sin }^{ - 1 }2t\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y={\cos }^{ - 1 }2t\text{\hspace{1} }\text{\hspace{1} }}$	解答

次のことを示せ．

$3$\displaystyle{z=f $ x,y $ }$ , $3$\displaystyle{x=r\cos {\theta}}$ , $3$\displaystyle{y=r\sin {\theta}}$ ならば
$3$\displaystyle{{ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ }^{2}+{ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{2}={ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}r \hspace{2}} $ }^{2}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} {r}^{2} \hspace{2}}{ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}{\theta} \hspace{2}} $ }^{2}}$ となる．解答

$3$\displaystyle{z=f $ x,y $ \text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }x=u\cos {\alpha}- v\sin {\alpha}\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=u\sin {\alpha}+v\cos {\alpha}\text{\hspace{1} }\text{\hspace{1} }}$ ならば，
$3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }{ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ }^{2}+{ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{2}={ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}u \hspace{2}} $ }^{2}+{ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}v \hspace{2}} $ }^{2}\text{\hspace{1} }\text{\hspace{1} }}$ となる．解答

次の関数の第2次偏導関数を求めよ．

$3$\displaystyle{z=3x- 2y}$	解答	$3$\displaystyle{z={x}^{3}+{y}^{3}}$	解答
$3$\displaystyle{z=2{x}^{2}- 3xy+4{y}^{2}}$	解答	$3$\displaystyle{z=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$	解答
$3$\displaystyle{z=\frac{\hspace{2} x- y \hspace{2}}{\hspace{2} x+y \hspace{2}}}$	解答	$3$\displaystyle{z=\sqrt{ 2x- 3y }}$	解答
$3$\displaystyle{z={e}^{ xy }}$	解答	$3$\displaystyle{z=\log $ x- y $ }$	解答
$3$\displaystyle{z=\sin \sqrt{ xy }}$	解答	$3$\displaystyle{z={\tan }^{ - 1 }\text{\hspace{1} }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$	解答

次の関数の第2次偏導関数を求めよ．

$3$\displaystyle{z=f $ x,y $ \text{\hspace{1} },}$
$3$\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=t- \sin t\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=1- \cos t\text{\hspace{1} }\text{\hspace{1} }}$ 解答
$3$\displaystyle{z=f $ x,y $ \text{\hspace{1} },}$
$3$\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=2{t}^{2}- 3\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y={t}^{2}+3t+7\text{\hspace{1} }\text{\hspace{1} }}$ 解答
次のことを示せ．

$3$\displaystyle{z=f $ x,y $ \text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }x=u\cos {\theta}- v\sin {\theta}\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=u\sin {\theta}+v\cos {\theta}\text{\hspace{1} }\text{\hspace{1} }}$ のとき，
$3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}=\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{v}^{2} \hspace{2}}\text{\hspace{1} }\text{\hspace{1} }}$ となる．解答

$3$\displaystyle{z=f $ x,y $ \text{\hspace{1}},\text{\hspace{1}}\text{\hspace{1}}x=r\cos {\theta}\text{\hspace{1}},\text{\hspace{1}}\text{\hspace{1}}y=r\sin {\theta}\text{\hspace{1}}\text{\hspace{1}}}$ のとき，
$3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}=\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{r}^{2} \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}r\hspace{2}}\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}r \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} {r}^{2} \hspace{2}}\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{{\theta}}^{2} \hspace{2}}\text{\hspace{1}}\text{\hspace{1}}}$ となる．解答

次のことを示せ．

$3$\displaystyle{y=f $ x,t $ }$ において，1次元の波動方程式

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}y \hspace{2}}{\hspace{2} {\partial}{t}^{2} \hspace{2}}={c}^{2}\frac{\hspace{2} {{\partial}}^{2}y \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}}$

に $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }{\xi}=x- ct\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }{\eta}=x+ct\text{\hspace{1} }\text{\hspace{1} }}$ なる変換を行うと

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}y \hspace{2}}{\hspace{2} {\partial}{\eta}{\partial}{\xi} \hspace{2}}=0}$

となる．

解答

$3$\displaystyle{z}$ に関する方程式

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{t}^{2} \hspace{2}}={c}^{2} $ \frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{r}^{2} \hspace{2}}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}r\hspace{2}}\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}r \hspace{2}} $ }$

に， $3$\displaystyle{z=\frac{\hspace{2}1\hspace{2}}{\hspace{2}r\hspace{2}}u}$ と置いて， $3$\displaystyle{u}$ の方程式とすると

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}u \hspace{2}}{\hspace{2} {\partial}{t}^{2} \hspace{2}}={c}^{2}\frac{\hspace{2} {{\partial}}^{2}u \hspace{2}}{\hspace{2} {\partial}{r}^{2} \hspace{2}}}$

となる

解答

次の関係で定義される陰関数 $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y={\varphi} $x$ \text{\hspace{1} }\text{\hspace{1} }}$ について $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}\text{\hspace{1} }\text{\hspace{1} }}$ を求めよ．

$3$\displaystyle{3{x}^{2}+2xy+{y}^{2}=1}$	解答	$3$\displaystyle{{y}^{2}=4px}$	解答
$3$\displaystyle{\frac{\hspace{2} {x}^{2} \hspace{2}}{\hspace{2} {a}^{2} \hspace{2}}+\frac{\hspace{2} {y}^{2} \hspace{2}}{\hspace{2} {b}^{2} \hspace{2}}=1}$	解答	$3$\displaystyle{y={e}^{ x+y }}$	解答
$3$\displaystyle{\log \text{\hspace{1} }\sqrt{ {x}^{2}+{y}^{2} }- {\tan }^{ - 1 }\text{\hspace{1} }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}=0}$	解答	$3$\displaystyle{{x}^{3}+{y}^{3}- 3axy=0}$	解答

次の関係で定義される陰関数 $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y={\varphi} $x$ \text{\hspace{1} }\text{\hspace{1} }}$ の指定された点における接線の方程式を求めよ．

$3$\displaystyle{\text{\hspace{1} \hspace{1} }f $ x,y $ =0\text{\hspace{1} }\text{\hspace{1} }}$ の点 $3$\displaystyle{\text{\hspace{1} } $ a,b $ \text{\hspace{1} }\text{\hspace{1} }}$ での接線の方程式解答

$3$\displaystyle{\text{\hspace{1} \hspace{1} }{x}^{3}+{y}^{3}- xy=0\text{\hspace{1} \hspace{1} }}$ の点 $3$\displaystyle{\text{\hspace{1} } $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}},\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \text{\hspace{1} }\text{\hspace{1} }}$ における接線の方程式解答

次の関係で定義される陰関数 $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y={\varphi} $x$ \text{\hspace{1} }\text{\hspace{1} }}$ の極値を調べよ．

$3$\displaystyle{{x}^{2}- 2xy+3{y}^{2}=8}$	解答	$3$\displaystyle{{x}^{2}y- x{y}^{2}+128=0}$	解答
$3$\displaystyle{{x}^{2}{y}^{2}- 2x+9{y}^{2}=0}$	解答	$3$\displaystyle{{x}^{3}- 12xy+2{y}^{3}=0}$	解答
$3$\displaystyle{{x}^{4}- 16xy+3{y}^{4}=0}$	解答	$3$\displaystyle{{x}^{4}+4{x}^{2}+3{y}^{3}- 2y=0}$	解答
$3$\displaystyle{{x}^{2}+2xy+{y}^{4}+2{y}^{2}=6}$	解答

次のことを示せ．

$3$\displaystyle{z=xf $ ax+by $ +yg $ ax+by $ \text{\hspace{1} }}$ ならば，
$3$\displaystyle{{b}^{2}\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}- 2ab\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}+{a}^{2}\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}=0}$

である
解答
次の関数をテイラー展開せよ．

$3$\displaystyle{f $ x,y $ ={x}^{2}+2xy+3{y}^{2}\text{\hspace{1} }}$ を点 $3$\displaystyle{\text{\hspace{1} } $ a,b $ \text{\hspace{1} }}$ でテイラー展開せよ．解答

次の関数の極値を求めよ．

$3$\displaystyle{f $ x,y $ ={x}^{2}+2{y}^{2}+10x}$	解答	$3$\displaystyle{f $ x,y $ ={x}^{2}- 2xy+3{y}^{2}- 4x+5y}$	解答
$3$\displaystyle{f $ x,y $ =4{x}^{2}+2xy+{y}^{2}+4x+4y}$	解答	$3$\displaystyle{f $ x,y $ =xy $ x- 2y- 3 $ }$	解答
$3$\displaystyle{f $ x,y $ ={x}^{3}+{y}^{3}- 12x- 27y}$	解答	$3$\displaystyle{f $ x,y $ ={x}^{3}+{y}^{3}+6xy- 24}$	解答
$3$\displaystyle{f $ x,y $ =xy+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}y\hspace{2}}}$	解答	$3$\displaystyle{f $ x,y $ =\cos x+\cos y+\cos $ x+y $ }$ $3$\displaystyle{\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} } $ 0< x\text{\hspace{1} },\text{\hspace{1} }y< {\pi} $ }$	解答

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