数学 A2 偏微分
問題 1 (復習). 指数で表せ.
1
(1) = x−1
x
(2)
1
= x−2
x2
(7)
√
3
x3 = x 2
(
(7)
2
1
(9) √ = x− 3
x3
(2) (x2 )x = 2x
(3) (x3 )x = 3x
y
2
= − 3
y
1
√
2 y
(9) (sin y)y = cos y
(5) (−5x3 )x = −15x
( )
1
1
(6)
= − 2
x x
x
問題 4. 偏導関数を求めよ
(1) z = 2x + y
{
2
= − 3
x
(2) z = 3x − 5y
{
1
√
2 x
(9) (sin x)x = cos x
(10) (cos x)x = − sin x
zx = 2
zy = 1
zx = 3
zy = −5
(3) z = −3x + 4y
{
zx = −3
zy = 4
zx = 4x − 1
zy = 6y
(7) z = 3x2 + xy − y 2
{
zx = 6x + y
zy = x − 2y
(8) z = −x2 + 2xy − 5y 2
{
zx = −2x + 2y
zy = 2x − 10y
(9) z = x2 + 2xy − y 2 − x + 3y
{
(10) (cos y)y = − sin y
1
(12) (log y)y =
y
(4) (3x2 )x = 6x
√
(8) ( x)x =
)
(11) (ey )y = ey
2
x
1
y2
√
(8) ( y)y =
問題 2. x について偏微分せよ.
(1) (x)x = 1
(7)
{
( )
1
1
(6)
= − 2
y y
y
1
1
(8) √ = x− 2
x
zx = 6x
zy = −2y
(6) z = 2x2 + 3y 2 − x
(5) (−5y 3 )y = −15y
(6)
)
{
(4) (3y )y = 6y
1
3
zx = 2x
zy = 4y
(5) z = 3x2 − y 2 + 5
2
√
3
1
x2
1
(12) (log x)x =
x
(3) (y 3 )y = 3y 2
1
√
(5) x = x 2
(
{
(2) (y 2 )y = 2y
1
= x−4
x4
x= x
(4) z = x2 + 2y 2
問題 3. y について偏微分せよ.
(1) (y)y = 1
1
(3) 3 = x−3
x
(4)
(11) (ex )x = ex
zx = 2x + 2y − 1
zy = 2x − 2y + 3
(10) z = 5x2 −xy+7y 2 +9x−4y
{
zx = 10x − y + 9
zy = −x + 14y − 4
y2
 x2
(11) z =

2y 2

 zx = − 3
x

2y

 zy = 2
x
√
(12) z = x3 y

√
z
=
3x
y

x

x3

 zy = √
2 y
数学 A2 偏微分
問題 1 (復習). 指数で表せ.
1
(1) = x−1
x
(2)
1
= x−2
x2
(7)
√
3
x3 = x 2
(
(7)
3
1
(9) √ = x− 2
x3
(3) (x )x = 3x
1
√
2 y
(9) (sin y)y = cos y
1
(12) (log y)y =
y
(4) (3x2 )x = 6x
(5) (−5x3 )x = −15x2
( )
1
1
= − 2
(6)
x x
x
問題 4. 偏導関数を求めよ
(1) z = 2x + y
{
2
= − 3
x
√
(8) ( x)x =
y
2
= − 3
y
(11) (ey )y = ey
2
x
)
(2) z = 3x − 5y
{
1
√
2 x
(9) (sin x)x = cos x
(10) (cos x)x = − sin x
zx = 2
zy = 1
zx = 3
zy = −5
(3) z = −3x + 4y
{
zx = −3
zy = 4
zx = 4x − 1
zy = 6y
(7) z = 3x2 + xy − y 2
{
zx = 6x + y
zy = x − 2y
(8) z = −x2 + 2xy − 5y 2
{
zx = −2x + 2y
zy = 2x − 10y
(9) z = x2 + 2xy − y 2 − x + 3y
{
(10) (cos y)y = − sin y
(2) (x2 )x = 2x
(7)
1
y2
√
(8) ( y)y =
問題 2. x について偏微分せよ.
(1) (x)x = 1
)
{
( )
1
1
= − 2
(6)
y y
y
1
1
(8) √ = x− 2
x
zx = 6x
zy = −2y
(6) z = 2x2 + 3y 2 − x
(5) (−5y 3 )y = −15y 2
(6)
1
x2
{
(4) (3y )y = 6y
1
3
zx = 2x
zy = 4y
(5) z = 3x2 − y 2 + 5
2
√
3
(
1
(12) (log x)x =
x
(3) (y 3 )y = 3y 2
1
√
(5) x = x 2
3
{
(2) (y 2 )y = 2y
1
= x−4
x4
x= x
(4) z = x2 + 2y 2
問題 3. y について偏微分せよ.
(1) (y)y = 1
1
(3) 3 = x−3
x
(4)
(11) (ex )x = ex
zx = 2x + 2y − 1
zy = 2x − 2y + 3
(10) z = 5x2 −xy+7y 2 +9x−4y
{
zx = 10x − y + 9
zy = −x + 14y − 4
y2
 x2
(11) z =

2y 2

 zx = − 3
x

2y

 zy = 2
x
√
(12) z = x3 y

2√

 zx = 3x y
x3

 zy = √
2 y

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