El cálculu multivariable (o cálculu en delles variables) nun ye más que la estensión del cálculu infinitesimal a funciones angulares y vectoriales de delles variables.
Campo angular con dos variables
Funciones de Rn en Rm . Campos angulares y vectoriales
Vamos Formular les definiciones pa campos vectoriales. Tamién van ser válides para campos angulares. Sía :
f
:
V
⟶
W
{\displaystyle \mathbf {f} :V\longrightarrow W}
P definíu biunívocamente pol so vector posición un vector
f
(
O
P
)
{\displaystyle \mathbf {f} {\big (}\mathbf {OP} {\big )}
O ye'l nuesu orixe de coordenaes .
V
⊆
R
n
,
W
⊆
R
m
,
{\displaystyle V\subseteq \mathbb {R} ^{n},W\subseteq \mathbb {R} ^{m},}
n
>
1
{\displaystyle n>1}
m
⩾
1
{\displaystyle m\geqslant 1}
m
=
1
{\displaystyle m=1}
campu angular . Pa
m
>
1
{\displaystyle m>1}
campu vectorial . Vamos Utilizar la norma euclídea pa topar la magnitú de los vectores .
Sean
a
∈
R
n
{\displaystyle \mathbf {a} \in \mathbb {R} ^{n}
b
∈
R
m
.
{\displaystyle \mathbf {b} \in \mathbb {R} ^{m}.}
lim
x
→
a
f
(
x
)
=
b
{\displaystyle \lim _{\mathbf {x} \to \mathbf {a} }\mathbf {f} {\big (}\mathbf {x} {\big )}=\mathbf {b} }
o bien, :
f
(
x
)
→
b
{\displaystyle \mathbf {f} (\mathbf {x} )\rightarrow \mathbf {b} }
x
→
a
{\displaystyle \mathbf {x} \rightarrow \mathbf {a} }
pa espresar lo siguiente:
lim
‖
x
−
a
‖
→
0
‖
f
(
x
)
−
b
‖
=
0
{\displaystyle \lim _{\big \|}\mathbf {x-a} {\big \|}\to 0}{\big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\big \|}=0}
onde
‖
x
‖
{\displaystyle {\big \|}\mathbf {x} {\big \|}
x
{\displaystyle \mathbf {x} }
x
=
(
x
1
,
…
,
x
n
)
,
a
=
(
a
1
,
…
,
a
n
)
,
{\displaystyle \mathbf {x} ={\big (}x_{1},\ldots ,x_{n}{\big )},\mathbf {a} ={\big (}a_{1},\ldots ,a_{n}{\big )},}
lim
(
x
1
,
…
,
x
n
)
→
(
a
1
,
…
,
a
n
)
f
(
x
1
,
…
,
x
n
)
=
b
{\displaystyle \lim _{\big (}x_{1},\ldots ,x_{n}{\big )}\to {\big (}a_{1},\ldots ,a_{n}{\big )}\mathbf {f} {\big (}x_{1},\ldots ,x_{n}{\big )}=\mathbf {b} }
o, de forma equivalente, :
lim
x
→
a
f
(
x
)
=
b
{\displaystyle \lim _{\mathbf {x} \to \mathbf {a} }\mathbf {f} {\big (}\mathbf {x} {\big )}=\mathbf {b} }
Dicimos qu'una función
f
{\displaystyle \mathbf {f} }
continua en
a
⇔
lim
x
→
a
f
(
x
)
=
f
(
a
)
{\displaystyle \mathbf {a} \Leftrightarrow \lim _{\mathbf {x} \to \mathbf {a} }\mathbf {f} {\big (}\mathbf {x} {\big )}=\mathbf {f} {\big (}\mathbf {a} {\big )}
Demostración
Sabemos qu'a) y b) nel teorema verifíquense si
f
{\displaystyle f}
g
{\displaystyle g}
b
=
(
b
1
,
…
,
b
m
)
,
c
=
(
c
1
,
…
,
c
m
)
{\displaystyle \mathbf {b} ={\big (}b_{1},\ldots ,b_{m}{\big )},\mathbf {c} ={\big (}c_{1},\ldots ,c_{m}{\big )}
a
)
f
(
x
)
=
[
f
1
(
x
)
,
…
,
f
m
(
x
)
]
,
g
(
x
)
=
[
g
1
(
x
)
,
…
,
g
m
(
x
)
]
lim
x
→
a
(
f
+
g
)
(
x
)
=
lim
x
→
a
[
(
f
1
+
g
1
)
(
x
)
,
…
,
(
f
m
+
g
m
)
(
x
)
]
=
[
lim
x
→
a
(
f
1
+
g
1
)
(
x
)
,
…
,
lim
x
→
a
(
f
m
+
g
m
)
(
x
)
]
=
[
lim
x
→
a
f
1
(
x
)
+
lim
x
→
a
g
1
(
x
)
,
…
,
lim
x
→
a
f
m
(
x
)
+
lim
x
→
a
g
m
(
x
)
]
=
(
b
1
+
c
1
,
…
,
b
m
+
c
m
)
=
(
b
1
,
…
,
b
m
)
+
(
c
1
,
…
,
c
m
)
=
b
+
c
{\displaystyle {\begin{array}{rl}a)&\mathbf {f} {\big (}\mathbf {x} )={\big [}f_{1}{\big (}\mathbf {x} {\big )},\ldots ,f_{m}{\big (}\mathbf {x} {\big )}{\big ]},\mathbf {g} {\big (}\mathbf {x} )={\Big [}g_{1}{\big (}\mathbf {x} {\big )},\ldots ,g_{m}{\big (}\mathbf {x} {\big )}{\Big ]}\\&\lim _{\mathbf {x} \to \mathbf {a} }{\big (}\mathbf {f} +\mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}=\lim _{\mathbf {x} \to \mathbf {a} }{\Big [}{\big (}f_{1}+g_{1}{\big )}{\big (}\mathbf {x} {\big )},\ldots ,{\big (}f_{m}+g_{m}{\big )}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\Big [}\lim _{\mathbf {x} \to \mathbf {a} }{\big (}f_{1}+g_{1}{\big )}{\big (}\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }{\big (}f_{m}+g_{m}{\big )}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\Big [}\lim _{\mathbf {x} \to \mathbf {a} }f_{1}{\big (}\mathbf {x} {\big )}+\lim _{\mathbf {x} \to \mathbf {a} }g_{1}(\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }f_{m}{\big (}\mathbf {x} {\big )}+\lim _{\mathbf {x} \to \mathbf {a} }g_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\big (}b_{1}+c_{1},\ldots ,b_{m}+c_{m}{\big )}={\big (}b_{1},\ldots ,b_{m}{\big )}+{\big (}c_{1},\ldots ,c_{m}{\big )}=\mathbf {b} +\mathbf {c} \end{array}
b
)
lim
x
→
a
λ
f
(
x
)
=
lim
x
→
a
λ
[
f
1
(
x
)
,
…
,
f
m
(
x
)
]
=
lim
x
→
a
[
λ
f
1
(
x
)
,
…
,
λ
f
m
(
x
)
]
=
[
lim
x
→
a
λ
f
1
(
x
)
,
…
,
lim
x
→
a
λ
f
m
(
x
)
]
=
[
λ
lim
x
→
a
f
1
(
x
)
,
…
,
λ
lim
x
→
a
f
m
(
x
)
]
=
λ
[
lim
x
→
a
f
1
(
x
)
,
…
,
lim
x
→
a
f
m
(
x
)
]
=
λ
(
b
1
,
…
,
b
m
)
=
λ
b
{\displaystyle {\begin{array}{rl}b)&\lim _{\mathbf {x} \to \mathbf {a} }\lambda \mathbf {f} {\big (}\mathbf {x} {\big )}=\lim _{\mathbf {x} \to \mathbf {a} }\lambda {\Big [}f_{1}{\big (}\mathbf {x} {\big )},\ldots ,f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\lim _{\mathbf {x} \to \mathbf {a} }{\Big [}\lambda f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lambda f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\Big [}\lim _{\mathbf {x} \to \mathbf {a} }\lambda f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }\lambda f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}={\Big [}\lambda \lim _{\mathbf {x} \to \mathbf {a} }f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lambda \lim _{\mathbf {x} \to \mathbf {a} }f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&\lambda {\Big [}\lim _{\mathbf {x} \to \mathbf {a} }f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\lambda {\big (}b_{1},\ldots ,b_{m}{\big )}=\lambda \mathbf {b} \end{array}
c
)
(
f
⋅
g
)
(
x
)
−
b
⋅
c
=
[
f
(
x
)
−
b
]
⋅
[
g
(
x
)
−
c
]
+
b
⋅
[
g
(
x
)
−
c
]
+
c
⋅
[
f
(
x
)
−
b
]
{\displaystyle c)\quad {\big (}\mathbf {f} \cdot \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}-\mathbf {b} \cdot \mathbf {c} ={\Big [}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big ]}\cdot {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big ]}+\mathbf {b} \cdot {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big ]}+\mathbf {c} \cdot {\Big [}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big ]}
Aplicando la desigualdá triangular y la desigualdá de Cauchy-Schwarz tenemos
|
(
f
⋅
g
)
(
x
)
−
b
⋅
c
|
⩽
‖
f
(
x
)
−
b
‖
⋅
‖
g
(
x
)
−
c
‖
+
‖
b
‖
⋅
‖
g
(
x
)
−
c
‖
+
‖
c
‖
⋅
‖
f
(
x
)
−
b
‖
⇒
0
⩽
lim
‖
x
−
a
‖
→
0
|
(
f
⋅
g
)
(
x
)
−
b
⋅
c
|
⩽
lim
‖
x
−
a
‖
→
0
‖
f
(
x
)
−
b
‖
⋅
lim
‖
x
−
a
‖
→
0
‖
g
(
x
)
−
c
‖
+
‖
b
‖
⋅
lim
‖
x
−
a
‖
→
0
‖
g
(
x
)
−
c
‖
+
‖
c
‖
lim
‖
x
−
a
‖
→
0
‖
f
(
x
)
−
b
‖
=
0
⋅
0
+
‖
b
‖
⋅
0
+
‖
c
‖
⋅
0
=
0
{\displaystyle {\begin{array}{l}{\Big |}{\big (}\mathbf {f} \cdot \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}-\mathbf {b} \cdot \mathbf {c} {\Big |}\leqslant {\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big \|}\cdot {\Big \|}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big \|}+{\big \|}\mathbf {b} {\big \|}\cdot {\Big \|}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big \|}+{\big \|}\mathbf {c} {\big \|}\cdot {\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big \|}\Rightarrow \\0\leqslant \lim _{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big |}{\big (}\mathbf {f} \cdot \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}-\mathbf {b} \cdot \mathbf {c} {\Big |}\leqslant \lim _{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big \|}\cdot \lim _{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big \|}+\\{\big \|}\mathbf {b} {\big \|}\cdot \lim _{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big \|}+{\big \|}\mathbf {c} {\big \|}\lim _{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big \|}=0\cdot 0+{\big \|}\mathbf {b} {\big \|}\cdot 0+{\big \|}\mathbf {c} {\big \|}\cdot 0=\\0\end{array}
, como queríamos demostrar.
d
)
g
(
x
)
=
f
(
x
)
,
c
=
b
⇒
lim
x
→
a
‖
f
(
x
)
‖
2
=
‖
b
‖
2
{\displaystyle d)\quad \mathbf {g} {\big (}\mathbf {x} {\big )}=\mathbf {f} {\big (}\mathbf {x} {\big )},\mathbf {c} =\mathbf {b} \Rightarrow \lim _{\mathbf {x} \to \mathbf {a} }{\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}{\Big \|}^{2}={\big \|}\mathbf {b} {\big \|}^{2}
Demostración
Sean
y
=
g
(
x
)
{\displaystyle \mathbf {y} =\mathbf {g} {\big (}\mathbf {x} {\big )}
b
=
g
(
a
)
{\displaystyle \mathbf {b} =\mathbf {g} {\big (}\mathbf {a} {\big )}
lim
‖
x
−
a
‖
→
0
‖
f
[
g
(
x
)
]
−
f
[
g
(
a
)
]
‖
=
lim
‖
y
−
b
‖
→
0
‖
f
(
y
)
−
f
(
b
)
‖
=
0
⇒
lim
x
→
a
f
[
g
(
x
)
]
=
f
[
g
(
a
)
]
{\displaystyle {\begin{array}{l}\lim _{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}-\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {a} {\big )}{\Big ]}{\Big \|}=\lim _{\big \|}\mathbf {y} -\mathbf {b} {\big \|}\to 0}{\Big \|}\mathbf {f} {\big (}\mathbf {y} {\big )}-\mathbf {f} {\big (}\mathbf {b} {\big )}{\Big \|}=0\Rightarrow \\\lim _{\mathbf {x} \to \mathbf {a} }\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}=\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {a} {\big )}{\Big ]}\end{array}
como queríamos demostrar.
Derivaes direccionales
Derivaes parciales
∂
f
∂
x
k
=
lim
h
→
0
f
(
x
1
,
…
,
x
k
+
h
,
…
,
x
n
)
−
f
(
x
1
,
…
,
x
k
,
…
,
x
n
)
h
{\displaystyle {\cfrac {\partial f}{\partial x_{k}=\lim _{h\to 0}{\cfrac {f{\big (}x_{1},\ldots ,x_{k}+h,\ldots ,x_{n}{\big )}-f{\big (}x_{1},\ldots ,x_{k},\ldots ,x_{n}{\big )}{h}
Si derivamos la espresión anterior al respective de una segunda variable,
x
j
{\displaystyle x_{j}
∂
2
f
∂
x
j
∂
x
k
{\displaystyle {\cfrac {\partial ^{2}f}{\partial x_{j}\partial x_{k}
∂
f
∂
x
k
{\displaystyle {\cfrac {\partial f}{\partial x_{k}
x
k
{\displaystyle x_{k}
x
j
,
∀
j
≠
k
{\displaystyle x_{j},\quad \forall j\neq k}
La diferencial
L'anterior ecuación ye la fórmula de Taylor de primer orde pa
f
(
a
+
v
)
{\displaystyle f{\big (}\mathbf {a} +\mathbf {v} {\big )}
Demostración
a
)
v
=
h
y
,
h
∈
R
,
lim
‖
v
‖
→
0
f
(
x
+
v
)
=
lim
‖
v
‖
→
0
f
(
x
+
h
y
)
=
f
(
x
)
+
f
L
(
h
y
)
=
f
(
x
)
+
h
f
L
(
y
)
⇒
lim
h
→
0
f
(
x
+
h
y
)
−
f
(
x
)
h
=
f
′
(
x
;
y
)
=
f
L
(
y
)
{\displaystyle {\begin{array}{rl}a)&\mathbf {v} =h\mathbf {y} ,\quad h\in \mathbb {R} ,\\&\lim _{\big \|}\mathbf {v} {\big \|}\to \mathbf {0} }f{\big (}\mathbf {x} +\mathbf {v} {\big )}=\lim _{\big \|}\mathbf {v} {\big \|}\to \mathbf {0} }f{\big (}\mathbf {x} +h\mathbf {y} {\big )}=f{\big (}\mathbf {x} {\big )}+f_{L}{\big (}h\mathbf {y} {\big )}=\\&f{\big (}\mathbf {x} {\big )}+hf_{L}{\big (}\mathbf {y} {\big )}\Rightarrow \\&\lim _{h\to 0}{\cfrac {f{\big (}\mathbf {x} +h\mathbf {y} {\big )}-f{\big (}\mathbf {x} {\big )}{h}=f'{\big (}\mathbf {x} ;\mathbf {y} {\big )}=f_{L}{\big (}\mathbf {y} {\big )}\end{array}
como queríamos demostrar.
b
)
{\displaystyle b)}
y
{\displaystyle y}
{
y
1
,
…
,
y
n
}
,
f
L
(
y
)
=
f
L
(
∑
k
=
1
n
y
k
y
k
)
=
∑
k
=
1
n
y
k
f
L
(
y
k
)
=
∑
k
=
1
n
y
k
f
′
(
x
;
y
k
)
=
∑
k
=
1
n
y
k
∂
f
∂
x
k
{\displaystyle {\begin{array}{l}{\big \{}\mathbf {y} _{1},\ldots ,\mathbf {y} _{n}{\big \},f_{L}{\big (}\mathbf {y} {\big )}=f_{L}{\big (}\sum _{k=1}^{n}y_{k}\mathbf {y} _{k}{\big )}=\sum _{k=1}^{n}y_{k}f_{L}{\big (}\mathbf {y} _{k}{\big )}=\sum _{k=1}^{n}y_{k}f'{\big (}\mathbf {x} ;\mathbf {y} _{k}{\big )}=\\\sum _{k=1}^{n}y_{k}{\cfrac {\partial f}{\partial x_{k}\end{array}
como queríamos demostrar.
Regla de la cadena
Espresando
f
′
(
x
;
y
)
{\displaystyle \mathbf {f'} {\big (}\mathbf {x} ;\mathbf {y} {\big )}
f
′
(
x
;
y
)
=
[
f
1
′
(
x
;
y
)
,
…
,
f
m
′
(
x
;
y
)
]
{\displaystyle \mathbf {f'} {\big (}\mathbf {x} ;\mathbf {y} {\big )}={\Big [}f'_{1}{\big (}\mathbf {x} ;\mathbf {y} {\big )},\ldots ,f'_{m}{\big (}\mathbf {x} ;\mathbf {y} {\big )}{\Big ]}
Esta ye la fórmula de Taylor de primer orde pa
f
.
f
L
(
v
)
=
f
′
(
x
;
v
)
{\displaystyle \mathbf {f} .\quad \mathbf {f} _{L}{\big (}\mathbf {v} {\big )}=\mathbf {f} '{\big (}\mathbf {x} ;\mathbf {v} {\big )}
La matriz de
f
′
{\displaystyle \mathbf {f} '}
Deduzse fácilmente de la fórmula de Taylor de primer orde yá vista.
Regla de la cadena pa diferenciales de campos vectoriales
∂
2
f
∂
x
i
∂
x
j
=
∂
2
f
∂
x
j
∂
x
i
∀
i
≠
j
⇔
{\displaystyle {\cfrac {\partial ^{2}f}{\partial x_{i}\partial x_{j}={\cfrac {\partial ^{2}f}{\partial x_{j}\partial x_{i}\quad \forall i\neq j\Leftrightarrow }
x
{\displaystyle \mathbf {x} }
Un campu angular tien un máximu en
x
=
a
⇔
{\displaystyle \mathbf {x} =\mathbf {a} \Leftrightarrow }
B
(
a
)
|
∀
x
∈
B
(
a
)
f
(
x
)
⩽
f
(
a
)
{\displaystyle B{\big (}\mathbf {a} {\big )}{\Big |}\forall \mathbf {x} \in B{\big (}\mathbf {a} {\big )}\quad f{\big (}\mathbf {x} {\big )}\leqslant f{\big (}\mathbf {a} {\big )}
Un campu angular tien un mínimu en
x
=
a
⇔
{\displaystyle \mathbf {x} =\mathbf {a} \Leftrightarrow }
B
(
a
)
|
∀
x
∈
B
(
a
)
f
(
x
)
⩾
f
(
a
)
{\displaystyle B{\big (}\mathbf {a} {\big )}{\Big |}\forall \mathbf {x} \in B{\big (}\mathbf {a} {\big )}\quad f{\big (}\mathbf {x} {\big )}\geqslant f{\big (}\mathbf {a} {\big )}
Un campu angular tien un puntu de ensilladura
⇔
{\displaystyle \Leftrightarrow }
∀
B
(
a
)
∃
x
|
f
(
x
)
⩽
f
(
a
)
∧
∃
x
|
f
(
x
)
⩾
f
(
a
)
{\displaystyle \forall B{\big (}\mathbf {a} {\big )}\quad \exists \mathbf {x} {\big |}f{\big (}\mathbf {x} {\big )}\leqslant f{\big (}\mathbf {a} {\big )}\land \exists \mathbf {x} {\big |}f{\big (}\mathbf {x} {\big )}\geqslant f{\big (}\mathbf {a} {\big )}
Función con un puntu de ensilladura Pa saber si ye unu de los casos anteriores:
Llogramos
x
|
∂
f
∂
x
k
=
0
∀
k
|
1
⩽
k
⩽
n
{\displaystyle \mathbf {x} {\Big |}{\cfrac {\partial f}{\partial x_{k}=0\qquad \forall k{\Big |}1\leqslant k\leqslant n}
Llogramos la matriz hessiana de f. Sía esta
F
(
x
)
{\displaystyle \mathbf {F} {\big (}\mathbf {x} {\big )}
F
(
x
)
{\displaystyle \mathbf {F} {\big (}\mathbf {x} {\big )}
⇒
f
{\displaystyle \Rightarrow f}
mínimu local (mínimu relativu ) en
x
{\displaystyle \mathbf {x} }
F
(
x
)
{\displaystyle \mathbf {F} {\big (}\mathbf {x} {\big )}
definida negativa
⇒
f
{\displaystyle \Rightarrow f}
máximu local (máximu relativu ) en
x
{\displaystyle \mathbf {x} }
F
(
x
)
{\displaystyle \mathbf {F} {\big (}\mathbf {x} {\big )}
indefinida
⇒
f
{\displaystyle \Rightarrow f}
puntu de ensilladura en
x
{\displaystyle \mathbf {x} }
No enantes espuesto, supunximos que
∂
2
f
∂
x
i
∂
x
j
{\displaystyle {\cfrac {\partial ^{2}f}{\partial x_{i}\partial x_{j}
∀
i
,
j
|
1
⩽
i
⩽
n
,
1
⩽
j
⩽
n
{\displaystyle \forall i,j{\big |}1\leqslant i\leqslant n,1\leqslant j\leqslant n}
Función diferenciable
Campu angular
Campu vectorial
Gradiente
Diverxencia
Rotacional
Integral de llinia
Integral de superficie
Integral múltiple Multiplicadores de Lagrange
Referencies
![{\displaystyle \lim _{\mathbf {x} \to \mathbf {a} }{\big [}\mathbf {f} +\mathbf {g} {\big ]}{\big (}\mathbf {x} {\big )}=\mathbf {b} +\mathbf {c} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/97e7d295bd1d5497f0c98c4d16c52d2720d749d0)
![{\displaystyle {\begin{array}{rl}a)&\mathbf {f} {\big (}\mathbf {x} )={\big [}f_{1}{\big (}\mathbf {x} {\big )},\ldots ,f_{m}{\big (}\mathbf {x} {\big )}{\big ]},\mathbf {g} {\big (}\mathbf {x} )={\Big [}g_{1}{\big (}\mathbf {x} {\big )},\ldots ,g_{m}{\big (}\mathbf {x} {\big )}{\Big ]}\\&\lim _{\mathbf {x} \to \mathbf {a} }{\big (}\mathbf {f} +\mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}=\lim _{\mathbf {x} \to \mathbf {a} }{\Big [}{\big (}f_{1}+g_{1}{\big )}{\big (}\mathbf {x} {\big )},\ldots ,{\big (}f_{m}+g_{m}{\big )}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\Big [}\lim _{\mathbf {x} \to \mathbf {a} }{\big (}f_{1}+g_{1}{\big )}{\big (}\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }{\big (}f_{m}+g_{m}{\big )}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\Big [}\lim _{\mathbf {x} \to \mathbf {a} }f_{1}{\big (}\mathbf {x} {\big )}+\lim _{\mathbf {x} \to \mathbf {a} }g_{1}(\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }f_{m}{\big (}\mathbf {x} {\big )}+\lim _{\mathbf {x} \to \mathbf {a} }g_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\big (}b_{1}+c_{1},\ldots ,b_{m}+c_{m}{\big )}={\big (}b_{1},\ldots ,b_{m}{\big )}+{\big (}c_{1},\ldots ,c_{m}{\big )}=\mathbf {b} +\mathbf {c} \end{array}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95faf263a1baf0db2c17757d92d4652b8000a3c0)
![{\displaystyle {\begin{array}{rl}b)&\lim _{\mathbf {x} \to \mathbf {a} }\lambda \mathbf {f} {\big (}\mathbf {x} {\big )}=\lim _{\mathbf {x} \to \mathbf {a} }\lambda {\Big [}f_{1}{\big (}\mathbf {x} {\big )},\ldots ,f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\lim _{\mathbf {x} \to \mathbf {a} }{\Big [}\lambda f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lambda f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\Big [}\lim _{\mathbf {x} \to \mathbf {a} }\lambda f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }\lambda f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}={\Big [}\lambda \lim _{\mathbf {x} \to \mathbf {a} }f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lambda \lim _{\mathbf {x} \to \mathbf {a} }f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&\lambda {\Big [}\lim _{\mathbf {x} \to \mathbf {a} }f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\lambda {\big (}b_{1},\ldots ,b_{m}{\big )}=\lambda \mathbf {b} \end{array}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10fef6b37ec01f2a38f73284854a3e4d6b80462c)
![{\displaystyle c)\quad {\big (}\mathbf {f} \cdot \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}-\mathbf {b} \cdot \mathbf {c} ={\Big [}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big ]}\cdot {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big ]}+\mathbf {b} \cdot {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big ]}+\mathbf {c} \cdot {\Big [}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66af9fd2140f9780543400554816ae1043fbe68f)
![{\displaystyle {\big (}\mathbf {f} \circ \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}=\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e050e088fa87cd14516180f245242f9c90b4d73)
![{\displaystyle {\begin{array}{l}\lim _{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}-\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {a} {\big )}{\Big ]}{\Big \|}=\lim _{\big \|}\mathbf {y} -\mathbf {b} {\big \|}\to 0}{\Big \|}\mathbf {f} {\big (}\mathbf {y} {\big )}-\mathbf {f} {\big (}\mathbf {b} {\big )}{\Big \|}=0\Rightarrow \\\lim _{\mathbf {x} \to \mathbf {a} }\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}=\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {a} {\big )}{\Big ]}\end{array}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e0972f84a181ff7e0e1de0688a2b517a3becc0b)
![{\displaystyle g(t)=f{\Big [}\mathbf {x} {\big (}t{\big )}{\Big ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c24d0a6ebe6f93545e22280032f94396d0320cf9)
![{\displaystyle \mathbf {f'} {\big (}\mathbf {x} ;\mathbf {y} {\big )}={\Big [}f'_{1}{\big (}\mathbf {x} ;\mathbf {y} {\big )},\ldots ,f'_{m}{\big (}\mathbf {x} ;\mathbf {y} {\big )}{\Big ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b886201a23ee2c88781d76fe6abf5585db77bc24)
![{\displaystyle \mathbf {h} '{\big (}\mathbf {x} {\big )}=\mathbf {f} '{\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}\circ \mathbf {g} '{\big (}\mathbf {x} {\big )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ee4c8d71f25c0b0392d15f41ec44fa1b485ee56)
