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 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^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))=-(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)

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