^{2}+y

^{3}-z.

Determine the Jacobian matrix Dz in P=(1,1,2) using implicit differentiation.

My idea is to calculate ∂z/∂x in P(1,1,2) and ∂z/∂y in P(1,1,2) and then just write it in matrix form.

So,

F(x,y,z)=x

^{2}+y

^{3}-z=0

∂z/∂x=-((∂F/∂x)/(∂F/∂z))=-(2x/-1)=2x

∂z/∂x in P(1,1,2)=2

∂z/∂y=-((∂F/∂y)/(∂F/∂z))=-(3y

^{2}/-1)= 3y

^{2}

∂z/∂y in P(1,1,2)=3

Dz (1,1,2) = (∂z/∂x(1,1,2) ∂z/∂y(1,1,2)), Dz is [1 X 2] matrix

Dz (1,1,2) = (2 3)

I checked the result using explicit differentiation and I obtained the same.

But in the book that I use I saw another approach. Namely, as a hint was given this formula:

D

_{x}=f(

__x__°)=-[D

_{y}F(

__x__°,

__y__°)]

^{-1}D

_{x}F(

__x__°,

__y__°)

I don’t understand how this formula can be used in order to calculate Dz.

Any help is appreciated.

Thanks in advance.