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April 8th, 2019, 10:42 AM   #1
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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.
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April 9th, 2019, 12:56 PM   #2
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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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