April 8th, 2019, 10:42 AM  #1 
Newbie Joined: Mar 2019 From:  Posts: 3 Thanks: 0  Implicit Differentiation
Given is the function F(x,y,z) = x^2+y^3z. 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^3z=0 ∂z/∂x=((∂F/∂x)/(∂F/∂z))=(2x/1)=2x ∂z/∂x in P(1,1,2)=2 ∂z/∂y=((∂F/∂y)/(∂F/∂z))=(3y2/1)= 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: Dx=f(x°)=[DyF(x°,y°)]1DxF(x°,y°) I don’t understand how this formula can be used in order to calculate Dz. Any help is appreciated. Thanks in advance. 
April 9th, 2019, 12:56 PM  #2 
Newbie Joined: Mar 2019 From:  Posts: 3 Thanks: 0 
It seems that both approaches are indeed equivalent. Since Dz is requested, Dx F(x°)=[Dy F(x°,y°)]^(1)Dx F(x°,y°) becomes DxyF(x°,y°)=[Dz F(x°,y°,z°)]^(1) Dxy F(x°,y°,z°) It follows that: [Dy F(x°,y°)]^(1)=[1]^(1) Dxy F(x°,y°,z°)=(2x 3x^2) Dxy F(1,1,2)=(2 3) So, Dz = [1]^(1) (2 3) = (2 3) 

Tags 
differentiation, implicit, implicit differentiation, implicit function theorem 
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