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```INVESTUGAR FUNCIONES DE VARIAS VARABLES
Si w=f(x,y,z) , entonces
𝜕𝑤
𝜕𝑦
= 𝑓𝑦 (𝑥, 𝑦, 𝑧) ;
𝜕𝑤
𝜕𝑧
𝜕𝑤
𝜕𝑥
= 𝑓𝑥 (𝑥, 𝑦, 𝑧)
= 𝑓𝑧 (𝑥, 𝑦, 𝑧)
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
(xy + yz + xz) =y+z→𝑓𝑥 (xy + yz + xz) =y+z
(xy + yz + xz) =x+z→𝑓𝑦 (xy + yz + xz) =x+z
(xy + yz + xz) =y+z→ 𝑓𝑧 (xy + yz + xz) = y+z
Si f(x,y,z) =xy+y𝑧 2 +xz, determine 𝑓𝑧 (x, y, z)
𝜕
𝜕𝑧
(xy + y𝑧 2 + xz)= 𝑓𝑧 (x, y, z) =2yz+x
a)f(x,y)=2x-5y+1
𝑓𝑥 (x,y)=2
𝑓𝑦 (x,y)=-5
b) f(x.y)=𝑥 2 − 3𝑦 2 + 4
𝑓𝑥 (𝑥, 𝑦)= 2x
𝑓𝑦 (𝑥, 𝑦)= -6y
c) z=x√𝑦
𝑓𝑥 (𝑥, 𝑦) =√𝑦
𝑓𝑦 (𝑥, 𝑦) =
𝑥
2√𝑦
d) z==𝑥 2 − 4𝑥𝑦 + 3𝑦 2 -5
𝑓𝑥 (𝑥, 𝑦) =2x-4y
𝑓𝑦 (𝑥, 𝑦) =-4x+6y
𝑑
𝑑𝑥
𝑒 𝑢 =u’𝑒 𝑢
e) z=𝑒 𝑥𝑦
𝑓𝑥 (𝑥, 𝑦)=y𝑒 𝑥𝑦
𝑓𝑦 (𝑥, 𝑦) = x𝑒 𝑥𝑦
f)z= 𝑥 2 𝑒 2𝑦
𝑓𝑥 (𝑥, 𝑦) = 2x𝑒 2𝑦
𝑓𝑦 (𝑥, 𝑦)= 2𝑥 2 𝑒 2𝑦
g) z=ln
c
𝑥
𝑦
1
𝑦
𝑥
𝑦
𝑓𝑥 (𝑥, 𝑦) = =
1
𝑥
−𝑥
𝑦2
𝑥
𝑦
𝑓𝑦 (𝑥, 𝑦) = = -
1
𝑦
h) z=ln(𝑥 2 𝑦 2 )
𝑓𝑥 (𝑥, 𝑦)=
𝑓𝑦 (𝑥, 𝑦)=
2𝑥𝑦 2
𝑥2𝑦2
=
2
𝑥
2𝑥 2 𝑦
2
𝑥2𝑦
𝑦
=
2
i) f(x,y,z)=sen(x+2y+3z)
𝑓𝑥 (𝑥, 𝑦, 𝑧) =cos(x+2y+3z)
𝑓𝑦 (𝑥, 𝑦, 𝑧) = 2cos(x+2y+3z)
𝑓𝑧 (𝑥, 𝑦, 𝑧) =3cos(x+2y+3z)
j) f(x,y,z)=3𝑥 2 𝑦 − 5𝑥𝑦𝑧 + 10𝑦𝑧 2
𝑓𝑥 (𝑥, 𝑦, 𝑧)=6xy-5yz
𝑓𝑦 (𝑥, 𝑦, 𝑧) =3𝑥 2 − 5𝑥𝑧+10𝑧 2
𝑓𝑧 (𝑥, 𝑦, 𝑧) =-5xy+20yz
a) f(x,y,z)=𝑥 3 𝑦𝑧 2 (1,1,1)
𝑓𝑥 (𝑥, 𝑦, 𝑧)=3𝑥 2 y𝑧 2 →𝑓𝑥 (𝑥, 𝑦, 𝑧)=3(1)2 (1)(1)2 =3
𝑓𝑦 (𝑥, 𝑦, 𝑧) =𝑥 3 𝑧 2 →𝑓𝑥 (𝑥, 𝑦, 𝑧)=(1)3 (1)2 =1
𝑓𝑧 (𝑥, 𝑦, 𝑧) =6𝑥 2 yz→𝑓𝑥 (𝑥, 𝑦, 𝑧)=2(1)3 (1)(1)=2
b) f(x,y,z)=𝑥 2 𝑦 3 + 2𝑥𝑦𝑧 − 3𝑦𝑧 (-2,1,2)
𝑓𝑥 (𝑥, 𝑦, 𝑧)=→𝑓𝑥 (𝑥, 𝑦, 𝑧)=2x𝑦 3 +2yz=2(-2)(1)3 +2(1)(2)=0
𝑓𝑦 (𝑥, 𝑦, 𝑧) =→𝑓𝑥 (𝑥, 𝑦, 𝑧) =3𝑥 2 𝑦 2 +2xz-3z=3(−2)2 (1)2 + 2(−2)(2)-3(2)=-2
𝑓𝑧 (𝑥, 𝑦, 𝑧) =→𝑓𝑥 (𝑥, 𝑦, 𝑧)=2xy-3y=2(-2)(1)-3(2)=-10
```