f(x, y)

微分積分 II
2014/10/10
1/3
問題
次の関数の偏導関数と二階偏導関数をすべて求めよ。
(i)
(iii)
√
f (x, y) = x + 2xy
f (x, y) = ln(x2 + y 2 )
(ii)
f (x, y) = 3x2 ey + xyex
解答
(i) 関数 f (x, y) を x に関して偏微分すると、
}
{
∂
∂ √
x + 2xy
f (x, y) =
∂x
∂x
{
}
∂
=
(x + 2xy)1/2
∂x
1
= (x + 2xy)1/2−1 × (1 + 2y)
2
1 + 2y
=
1
2(x + 2xy) 2
{
}
∂
∂
1/2
fy (x, y) =
f (x, y) =
(x + 2xy)
∂y
∂y
1
= (x + 2xy)1/2−1 × 2x
2
x
=
1
(x + 2xy) 2
fx (x, y) =
(1)
(2)
ここで、式 (1) を用いて
{
}
∂
∂
1 + 2y
fx (x, y) =
1
∂x
∂x 2(x + 2xy) 2
{
}
1
∂ 1 + 2y
=
(x + 2xy)− 2
∂x
2
{
}
1
1 + 2y
∂
=
×
(x + 2xy)− 2
2
∂x
{
}
1 + 2y
1
=
× − (x + 2xy)−1/2−1 × (1 + 2y)
2
2
(1 + 2y)2
(1 + 2y)2
=−
(x + 2xy)−5/2 = −
5
4
4(x + 2xy) 2
{
}
∂
∂
1 + 2y
fxy (x, y) =
fx (x, y) =
∂y
∂y 2(x + 2xy) 12
}
{
1 ∂
1 + 2y
=
1
2 ∂y (x + 2xy) 2
1
1
1/2−1
2
× 2x
1 2 × (x + 2xy) − (1 + 2y) × 2 (x + 2xy)
= ×
2
(x + 2xy)
x(1 + 2y)
1
=
1 −
3
(x + 2xy) 2
2(x + 2xy) 2
fxx (x, y) =
さらに、式 (2) を用いて
fyx (x, y) =
{
}
∂
∂
x
fy (x, y) =
∂x
∂x (x + 2xy) 12
1/3
微分積分 II
2014/10/10
2/3
1
=
1
2
−
x(1 + 2y)
1
(x + 2xy)
2(x + 2xy) 2
{
}
∂
∂
x
fyy (x, y) =
fy (x, y) =
∂y
∂y (x + 2xy) 23
=−
x2
1
(x + 2xy) 2
(ii)
{
}
∂
∂
f (x, y) =
3x2 ey + xyex
∂x
∂x
= 6xey + yex + xyex
{
}
∂
∂
fy (x, y) =
f (x, y) =
3x2 ey + xyex
∂y
∂y
fx (x, y) =
= 3x2 ey + xex
(3)
(4)
ここで、式 (3) を用いて
{
}
∂
∂
fx (x, y) =
6xey + yex + xyex
∂x
∂x
= 6ey + 1 × yex + yex + xyex
{
}
= 6ey + 2y + xy ex
{
}
∂
∂
y
x
fxy (x, y) =
fx (x, y) =
6xe + (1 + x)ye
∂y
∂y
y
= 6xe + (1 + x)ex
fxx (x, y) =
さらに、式 (4) を用いて
{
}
∂
∂
fy (x, y) =
3x2 ey + xex
∂x
∂x
{
}
= 6xey + ex + xex = 6xey + (1 + x)ex
{
}
∂
∂
fyy (x, y) =
fy (x, y) =
3x2 ey + xex
∂y
∂y
fyx (x, y) =
= 3x2 ey
(iii)
{
}
∂
∂
f (x, y) =
ln(x2 + y 2 )
∂x
∂x
2x
= 2
x + y2
{
}
∂
∂
2
2
fy (x, y) =
f (x, y) =
ln(x + y )
∂y
∂y
2y
= 2
x + y2
fx (x, y) =
(5)
(6)
ここで、式 (5) を用いて
fxx (x, y) =
{
}
∂
∂
2x
fx (x, y) =
∂x
∂x x2 + y 2
2 × (x2 + y 2 ) − 2x × 2x
=
(x2 + y 2 )2
2/3
2014/10/10
微分積分 II
3/3
2y 2 − 2x2
2(x2 + y 2 ) − 4x2
= 2
2
2
2
(x + y )
(x + y 2 )2
{
}
∂
2x
∂
fx (x, y) =
fxy (x, y) =
∂y
∂y x2 + y 2
2x × (2y)
4xy
=− 2
=− 2
(x + y 2 )2
(x + y 2 )2
=
さらに、式 (6) を用いて
{
}
∂
∂
2y
fy (x, y) =
fyx (x, y) =
∂x
∂x x2 + y 2
2y × 2x
4xy
=− 2
=− 2
(x + y 2 )2
(x + y 2 )2
{
}
∂
∂
2y
fy (x, y) =
fyy (x, y) =
∂y
∂y x2 + y 2
2 × (x2 + y 2 ) − 2y × 2y
=
(x2 + y 2 )2
2x2 − 2y 2
= 2
(x + y 2 )2
3/3

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