# Implicit Differentiation - Different Approaches

#### cargar

##### New member
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.

#### HallsofIvy

##### Elite Member
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

##### New member
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)

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#### cargar

##### New member
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)