x + y

問題 (1) fxx , fxy , fyy を計算せよ. (1) f (x, y) = sin(x cos y), (2) f (x, y) = log(x + ey ),
(3) (x + y)2 + (x − y)3
解 (1)
∂f
∂f
= cos(x cos y) · cos y,
= cos(x cos y) · (−x sin y),
∂x
∂y
∂2f
= − sin(x cos y) · (cos y)2 ,
∂x2
∂2f
= − sin(x cos y) · (−x sin y) · cos y + cos(x cos y) · (− sin y)
∂x∂y
= x sin y cos y sin(x cos y) − sin y cos(x cos y)
∂2f
= − sin(x cos y) · (−x sin y) · (−x sin y) + cos(x cos y) · (−x cos y)
∂y 2
= −x2 sin y cos y sin(x cos y) − x cos y cos(x cos y)
(2)
∂f
x
∂f
ey
=
,
=
,
∂x
x + ey ∂y
x + ey
∂2f
(x + ey ) − x
ey
=
=
,
∂x2
(x + ey )2
(x + ey )2
∂2f
−xey
=
∂x∂y
(x + ey )2
∂2f
ey (x + ey ) − ey · ey
xey
=
=
∂y 2
(x + ey )2
(x + ey )2
(3)
∂f
∂f
= 2(x + y) + 3(x − y)2 ,
= 2(x + y) − 3(x − y)2 ,
∂x
∂y
2
∂ f
= 2 + 6(x − y),
∂x2
∂2f
= 2 − 6(x − y)
∂x∂y
∂2f
= 2 + 6(x − y)
∂y 2

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