Multivariable calculus

Question

Consider that the system of three equations

xy²+zu+v²=3

x³z+2y-uv=2

xu+yv-xyz=1

defines implicitly x y and z as a function of u and v

Find ∂y/∂v and ∂z/∂u at values x=y=z=u=v=1

Answer 1

xy²+zu+v²=3

x³z+2y-uv=2

xu+yv-xyz=1

defines implicitly x y and z as a function of u and v

Find ∂y/∂v and ∂z/∂u at values

[∂/∂v] (xy²+zu+v²) =0 ⇒ 2x y ∂y/∂v + u ∂z/∂v +2v = 0 and at x=y=z=u=v=1 ⇒ ∂y/∂v + ∂z/∂v = -2 ( Eq. 1)

[∂/∂v]( x³z+2y-uv ) =0 ⇒3x² z ∂x/∂v +x³∂z/∂v + 2 ∂y/∂v - u = 0 and at x=y=z=u=v=1 ⇒

3∂x/∂v+∂z/∂v +2 ∂y/∂v =1 ( Eq. 2)

[∂/∂v] ( xu+yv-xyz )= 0 ⇒u∂x/∂v +y + v∂y/∂v +yz∂x/∂v + xz ∂y/∂v + xy ∂z/∂v = 0 and at x=y=z=u=v=1 ⇒

∂x/∂v +∂y/∂v +∂x/∂v + ∂y/∂v +∂z/∂v = 0 ⇒ 2∂x/∂v +2 ∂y/∂v +∂z/∂v = 0 ( Eq.3)

Now notice That Equations 1, 2, and 3 form a linear 3x3 system By letting ∂x/∂v = λ ,∂y/∂v= μ and ∂z/∂v =ξ

we have the 3x3 linear system.

μ + ξ = -2

3λ+2μ +ξ =1

2λ +2 μ +ξ = 0

Now the ball is in your court