Esercizi svolti di analisi reale e complessa - Dipartimento di ...

F ′ (t) =
∂ f
∂ x (g(t), 1 − g2 (t))g ′ ∂ f
(t) +
∂ y (g(t), 1 − g2 (t))(−2g(t)g ′ (t)) .
g(0) = 1
g ′ (t) =
1
cosh 2 t +
t
√
1 + t2 g ′ (0) = 1
1)
F ′ (0) =
=
∂ f
f
(1, 0) − 2∂ (1, 0) =
∂ x ∂ y
−2
.
0
∂
∂x
=
∂
∂z
=
f(x, h(x, z), z) = ∂f
∂x
∂ f
(x, h(x, z), z),
∂ x
f(x, h(x, z), z) = ∂f
∂z
∂ f
(x, h(x, z), z),
∂ z
(x, h(x, z), z) + ∂f
∂y
∂ f
(x, h(x, z), z),
∂ y
(x, h(x, z), z) + ∂f
∂y
∂ f
(x, h(x, z), z),
∂ y
∂h
(x, h(x, z), z) (x, z) =
∂x
∂ f
(x, h(x, z), z)
∂ z
∂h
(x, h(x, z), z) (x, z) =
∂z
∂ f
(x, h(x, z), z)
∂ z
⎛ ⎞
1
⎜ ∂ h ⎟
· ⎝ (x, z) ⎠
∂ x
0
⎛
⎜
· ⎝
y = f(x) C 2 x = 0
x 2 + sinh y + e xy = 1 ,
f(0) = 0
F (x) ≡ x 2 + sinh f(x) + e xf(x) − 1 ≡ 0
0
∂ h
(x, z)
∂ z
1
⎞
⎟
⎠ .