DERIVADA PARCIAL de funciones Reales Soluciones 2
∂ f ∂ f
1) f(x,y)= xy + x2y2 ; en (1,2) Por la definición.
∂x ∂y
∂ f f((1,2)+λ(1,0)) - f(1,2) f(1+λ,2) - f(1,2) f(1,2) = 6
a) (1,2)= lim = lim
∂x λ λ f(1+λ,2) = 4λ2+10λ+6
λ→0 λ→0
∂ f 4λ2+10λ+6 - 6 λ(4λ+10)
(1,2)= lim = lim = lim (4λ+10) = 10
∂x λ λ
λ→0 λ→0 λ→0
∂ f f((1,2)+λ(0,1)) - f(1,2) f(1,2+λ) - f(1,2) f(1,2) = 6
b) (1,2)= lim = lim
∂y λ λ f(1,2+λ) = λ2+5λ+6
λ→0 λ→0
∂ f λ2+5λ+6 - 6 λ(λ+5)
(1,2)= lim = lim = lim (λ+5) = 5
∂y λ λ
λ→0 λ→0 λ→0
∂ f ∂ f ∂ f
2) f(x,y,z)  = 2x2y3- xyz + 5z3 ; ; en (1,-1,1)
∂x ∂y ∂z
∂ f ∂ f ∂ f
a) = 4xy3- yz b) = 6x2y2- xz c) = - xy+10z
∂x ∂y ∂z
∂ f ∂ f ∂ f
(1,-1,1)= -3 (1,-1,1)= 5 (1,-1,1)= 16
∂x ∂y ∂z
∂ f ∂ f
3) f(x,y)= x2- 3xy ; en (-1,-1)
∂x ∂y
∂ f ∂ f
a) (-1,-1) = 2x - 3y b) (-1,-1) = - 3x
∂x ∂y
∂ f ∂ f
(-1,-1) = 1 (-1,-1) = 3
∂x ∂y
xy ∂ f ∂ f
4) f(x,y)= ; en (2,1)
x2- y2 ∂x ∂y
∂ f y3- x2y ∂ f x3+xy2
a) = b) =
∂x (x2- y2)2 ∂y (x2- y2)2
∂ f ∂ f
(2,1) = -1/3 (2,1) = 10/9
∂x ∂y