1.1.67

67.- Encuentre todas las segundas derivadas parciales de

f
(
x
,
y
)

sen

2

⁡
(
2
x
3
y
)

{\displaystyle f(x,y)\operatorname {sen} ^{2}(2x+3y)}

f
(
x
y
# )
sen
⁡
(
2

x

2

3
y
)

{\displaystyle f(xy)=\operatorname {sen}(2x^{2}+3y)}

Derivamos con respecto de x

2
(

∂

∂
x

sen
⁡
(
2
x
3
y
)
)
sen
⁡
(
2
x
3
y
)

{\displaystyle 2({\partial \over \partial x}\operatorname {sen}(2x+3y))\operatorname {sen}(2x+3y)}

(
cos
⁡
(
2
x
3
y
)
)
(

∂

∂
x

(
2
x
3
y
)
)
2
sen
⁡
(
2
x
3
y
)

{\displaystyle (\cos(2x+3y))({\partial \over \partial x}(2x+3y))2\operatorname {sen}(2x+3y)}

=
(
2
(

∂

∂
x

(
x
)
)

∂

∂
x

(
3
y
)
# )
2
cos
⁡
(
2
x
3
y
)
sen
⁡
(
2
x
3
y
)

{\displaystyle =(2({\partial \over \partial x}(x))+{\partial \over \partial x}(3y))=2\cos(2x+3y)\operatorname {sen}(2x+3y)}

=
2
cos
⁡
(
2
x
3
y
)
sen
⁡
(
2
x
3
y
)
(

∂

∂
x

2
x
(
3
y
)
# )
2
cos
⁡
(
2
x
3
y
)
sen
⁡
(
2
x
3
y
)
(
2
)

{\displaystyle =2\cos(2x+3y)\operatorname {sen}(2x+3y)({\partial \over \partial x}2x+(3y))=2\cos(2x+3y)\operatorname {sen}(2x+3y)(2)}

Lo que nos da:

=
4
cos
⁡
(
2
x
3
y
)
sen
⁡
(
2
x
3
y
)

{\displaystyle =4\cos(2x+3y)\operatorname {sen}(2x+3y)}

Derivamos de nuevo esa función con respecto de x:

4
(

∂

∂
x

(
c
o
s
(
2
x
3
y
)
s
e
n
(
2
x
3
y
)
)
)

{\displaystyle 4 ({\partial\over \partial x}(cos(2x+3y)sen(2x+3y)))}

=
4
(
c
o
s
(
2
x
3
y
)
(

∂

∂
x

(
s
e
n
(
2
x
3
y
)
)
)
(

∂

∂
x

(
c
o
s
(
2
x
3
y
)
)
)
s
e
n
(
2
x
3
y
)
)

{\displaystyle =4 (cos(2x+3y)({\partial\over \partial x}(sen(2x+3y)))+({\partial\over \partial x}(cos(2x+3y)))sen(2x+3y))}

=
4
(
c
o

s
2

(
2
x
3
y
)
(

∂

∂
x

(
2
x
3
y
)
)
(

∂

∂
x

(
c
o
s
(
2
x
3
y
)
)
)
s
e
n
(
2
x
3
y
)
)

{\displaystyle =4(cos^2(2x+3y)({\partial\over \partial x}(2x+3y))+({\partial\over \partial x}(cos(2x+3y)))sen(2x+3y))}

=
4
(
2
c
o

s
2

(
2
x
3
y
)
(

∂

∂
x

(
c
o
s
(
2
x
3
y
)
)
)
s
e
n
(
2
x
3
y
)
)

{\displaystyle = 4(2cos^2(2x+3y)+({\partial\over \partial x}(cos(2x+3y)))sen(2x+3y))}

=
4
(
2
c
o

s
2

(
2
x
3
y
)
−
(

∂

∂
x

(
2
x
3
y
)
)
s
e

n
2

(
2
x
3
y
)

{\displaystyle =4(2cos^2(2x+3y)-({\partial\over \partial x}(2x+3y))sen^2(2x+3y)}

=
4
(
2
c
o

s
2

(
2
x
3
y
)
−
2
s
e

n
2

(
2
x
3
y
)
)

{\displaystyle =4(2cos^2(2x+3y)-2sen^2(2x+3y))}

Lo que nos da:

f
2

(
x
x
# )
8
(
c
o

s
2

(
2
x
3
y
)
−
s
e

n
2

(
2
x
3
y
)
)

{\displaystyle f^2(xx) = 8(cos^2(2x+3y)- sen^2(2x+3y))}

Derivamos

4
cos
⁡
(
2
x
3
y
)
sen
⁡
(
2
x
3
y
)

{\displaystyle 4\cos(2x+3y)\operatorname {sen}(2x+3y)}

 de nuevo pero ahora con respecto de y

4
(

∂

∂
y

(
cos
⁡
(
2
x
3
y
)
sen
⁡
(
2
x
3
y
)
)
)

{\displaystyle 4({\partial \over \partial y}(\cos(2x+3y)\operatorname {sen}(2x+3y)))}

=
4
(
cos
⁡
(
2
x
3
y
)
(

∂

∂
y

(
sen
⁡
(
2
x
3
y
)
)
)
(

∂

∂
y

(
cos
⁡
(
2
x
3
y
)
)
(
sen
⁡
(
2
x
3
y
)
)

{\displaystyle =4(\cos(2x+3y)({\partial \over \partial y}(\operatorname {sen}(2x+3y)))+({\partial \over \partial y}(\cos(2x+3y))(\operatorname {sen}(2x+3y))}

=
4
(

cos

2

⁡
(
2
x
3
y
)
(

∂

∂
y

(
2
x
3
y
)
)
(

∂

∂
y

(
cos
⁡
(
2
x
3
y
)
)
(
sen
⁡
(
2
x
3
y
)
)

{\displaystyle =4(\cos ^{2}(2x+3y)({\partial \over \partial y}(2x+3y))+({\partial \over \partial y}(\cos(2x+3y))(\operatorname {sen}(2x+3y))}

=
4
(
3

cos

2

⁡
(
2
x
3
y
)
(

∂

∂
y

(
cos
⁡
(
2
x
3
y
)
)
(
sen
⁡
(
2
x
3
y
)
)

{\displaystyle =4(3\cos ^{2}(2x+3y)+({\partial \over \partial y}(\cos(2x+3y))(\operatorname {sen}(2x+3y))}

=
4
(
3

cos

2

⁡
(
2
x
3
y
)
(

∂

∂
y

(
2
x
3
y
)
(

sen

2

⁡
(
2
x
3
y
)
)

{\displaystyle =4(3\cos ^{2}(2x+3y)+({\partial \over \partial y}(2x+3y)(\operatorname {sen} ^{2}(2x+3y))}

=
4
(
3

cos

2

⁡
(
2
x
3
y
)
−
3

sen

2

⁡
(
2
x
3
y
)
)

{\displaystyle =4(3\cos ^{2}(2x+3y)-3\operatorname {sen} ^{2}(2x+3y))}

Lo que nos da:

=

f
2

(
x
y
# )
12

cos
2

⁡
(
2
x
3
y
)
−
s
e

n
2

(
2
x
3
y
)
)

{\displaystyle = f^2(xy) = 12 \cos^2 (2x+3y) - sen^2(2x+3y))}

Ahora derivamos la función original con respecto de y :

=
sen
⁡
(
2

x

2

3
y
)

{\displaystyle =\operatorname {sen}(2x^{2}+3y)}

=
2
(

∂

∂
y

(
sen
⁡
(
2
x
3
y
)
)
)
(
sen
⁡
(
2
x
3
y
)
)

{\displaystyle =2({\partial \over \partial y}(\operatorname {sen}(2x+3y)))(\operatorname {sen}(2x+3y))}

=
2
sen
⁡
(
2
x
3
y
)
(
cos
⁡
(
2
x
3
y
)
(

∂

∂
y

(
2
x
3
y
)
)
)

{\displaystyle =2\operatorname {sen}(2x+3y)(\cos(2x+3y)({\partial \over \partial y}(2x+3y)))}

=
2
sen
⁡
(
2
x
3
y
)
(
cos
⁡
(
2
x
3
y
)
(
3
(

∂

∂
y

(
y
)
)
)

{\displaystyle =2\operatorname {sen}(2x+3y)(\cos(2x+3y)(3({\partial \over \partial y}(y)))}

Que da lo siguiente:

=

6
sen
⁡
(
2
x
3
y
)
c
o
s
(
2
x
3
y
)

{\displaystyle 6\operatorname {sen}(2x+3y)cos(2x+3y)}

Esta nueva función la derivamos con respecto de x :

f

2

(
y
x
# )
6
sen
⁡
(
2
x
3
y
)
c
o
s
(
2
x
3
y
)

{\displaystyle f^{2}(yx)=6\operatorname {sen}(2x+3y)cos(2x+3y)}

=
6
(

∂

∂
x

(
sen
⁡
(
2
x
3
y
)
c
o
s
(
2
x
3
y
)
)
)

{\displaystyle =6({\partial \over \partial x}(\operatorname {sen}(2x+3y)cos(2x+3y)))}

=
6
(

∂

∂
x

(
sen
⁡
(
2
x
3
y
)
c
o
s
(
2
x
3
y
)
)
)
sen
⁡
(
2
x
3
y
)
(

∂

∂
x

(
c
o
s
(
2
x
3
y
)
)
)

{\displaystyle =6({\partial \over \partial x}(\operatorname {sen}(2x+3y)cos(2x+3y)))+\operatorname {sen}(2x+3y)({\partial \over \partial x}(cos(2x+3y)))}

=
6
(

cos

2

⁡
(
2
x
3
y
)
(

∂

∂
x

(
2
x
3
y
)
)
sen
⁡
(
2
x
3
y
)
(

∂

∂
x

(
c
o
s
(
2
x
3
y
)

{\displaystyle =6(\cos ^{2}(2x+3y)({\partial \over \partial x}(2x+3y))+\operatorname {sen}(2x+3y)({\partial \over \partial x}(cos(2x+3y)}

=
6
(
2

cos

2

⁡
(
2
x
3
y
)
sen
⁡
(
2
x
3
y
)
(

∂

∂
x

(
c
o
s
(
2
x
3
y
)
)
)

{\displaystyle =6(2\cos ^{2}(2x+3y)+\operatorname {sen}(2x+3y)({\partial \over \partial x}(cos(2x+3y)))}

=
6
(
2

cos

2

⁡
(
2
x
3
y
)
−
(

∂

∂
x

(
2
x
3
y
)
)

sen

2

⁡
(
2
x
3
y
)
)

{\displaystyle =6(2\cos ^{2}(2x+3y)-({\partial \over \partial x}(2x+3y))\operatorname {sen} ^{2}(2x+3y))}

=
6
(
2

cos

2

⁡
(
2
x
3
y
)
−
2

sen

2

⁡
(
2
x
3
y
)
)

{\displaystyle =6(2\cos ^{2}(2x+3y)-2\operatorname {sen} ^{2}(2x+3y))}

Lo que nos da:

=
12
(
c
o

s
2

(
2
x
3
y
)
−
s
e

n
2

(
2
x
3
y
)
)

{\displaystyle =12(cos^2(2x+3y)-sen^2(2x+3y))}

Ahora con respecto de y:

f

2

(
y
y
# )
6
sen
⁡
(
2
x
3
y
)
c
o
s
(
2
x
3
y
)

{\displaystyle f^{2}(yy)=6\operatorname {sen}(2x+3y)cos(2x+3y)}

=
6
(

∂

∂
y

sen
⁡
(
2
x
3
y
)
c
o
s
(
2
x
3
y
)
)

{\displaystyle =6({\partial \over \partial y}\operatorname {sen}(2x+3y)cos(2x+3y))}

=
6
(
sen
⁡
(
2
x
3
y
)
(

∂

∂
y

(
c
o
s
(
2
x
3
y
)
)
)
(

∂

∂
y

(
sen
⁡
(
2
x
3
y
)
)
)
c
o
s
(
2
x
3
y
)
)

{\displaystyle =6(\operatorname {sen}(2x+3y)({\partial \over \partial y}(cos(2x+3y)))+({\partial \over \partial y}(\operatorname {sen}(2x+3y)))cos(2x+3y))}

=
6
(

sen

2

⁡
(

∂

∂
y

(
2
x
3
y
)
)
(

∂

∂
y

(
2
x
3
y
)
)
c
o

s

2

(
2
x
3
y
)
)

{\displaystyle =6(\operatorname {sen} ^{2}({\partial \over \partial y}(2x+3y))+({\partial \over \partial y}(2x+3y))cos^{2}(2x+3y))}

=
6
(
3
c
o

s

2

(
2
x
3
y
)
−
3

sen

2

⁡
(
2
x
3
y
)
)

{\displaystyle =6(3cos^{2}(2x+3y)-3\operatorname {sen} ^{2}(2x+3y))}

Lo que nos da:

=
18
(
c
o

s
2

(
2
x
3
y
)
−
s
e

n
2

(
2
x
3
y
)
)

{\displaystyle =18 (cos^2 (2x+3y)-sen^2(2x+3y))}