Implicit Differentiation - Different Approaches

cargar

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Mar 19, 2019
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Given is the function F(x,y,z) = x2+y3-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)=x2+y3-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)= 3y2

∂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.
 

HallsofIvy

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Frankly, I don't understand the notation of the hint. But what you did I would not consider "implicit differentiation. Given \(\displaystyle F(x,y,z)= x^3+ y^2- z= 0\), assuming that x and y are independent variables and that z is a function of x and y then \(\displaystyle 3x^2- \frac{\partial z}{\partial x}= 0\) so \(\displaystyle \frac{\partial z}{\partial x}= 3x^2\). In fact, you could just write \(\displaystyle z= x^3+ y^2\) and use "regular" partial differentiation.
 

cargar

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But what you did I would not consider "implicit differentiation.
The above mentioned approach I found in the book Calculus (James Stewart).

Given \(\displaystyle F(x,y,z)= x^3+ y^2- z= 0\), assuming that ...
The original function was \(\displaystyle F(x,y,z)= x^2+ y^3- z= 0 \) and not \(\displaystyle F(x,y,z)= x^3+ y^2- z= 0 \)

In fact, you could just write \(\displaystyle z= x^3+ y^2\) and use "regular" partial differentiation.
But that would be explicit differentiation, wouldn't it?


It seems that both approaches are indeed equivalent.

Since Dz is requested, Dxf(x°)=-[DyF(x°,y°)]-1DxF(x°,y°) becomes Dxyf(x°,y°)=-[DzF(x°,y°,z°)]-1 DxyF(x°,y°,z°)


It follows that:

-[DyF(x°,y°)]-1=-[-1]-1

DxyF(x°,y°,z°)=(2x 3x2)


DxyF(1,1,2)=(2x 3x2)

So,

Dz = -[-1]-1 (2x 3x2) = (2 3)
 
Last edited:

cargar

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I cannot edit my post, but there is a typo Dxy F(1,1,2) = (2 3) not Dxy F(1,1,2) = (2x 3x2).

So, Dz = -[-1]-1 (2 3) = (2 3)
 
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