Implicit Differentiation - Different Approaches 2

Question

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 $\frac{∂z}{∂x}$ in $P(1,1,2)$ and $\frac{∂z}{∂y}$ in $P(1,1,2)$ and then just write it in matrix form.

So,

$F(x,y,z)=x^2+y^3-z=0$

$\frac{∂z}{∂x}=-\frac{\frac{∂F}{∂x}}{\frac{∂F}{∂z}}=-{\frac{2x}{-1}}=2x$

$\frac{∂z}{∂x} $ in $ P(1,1,2)=2$

$\frac{∂z}{∂y}=-\frac{\frac{∂F}{∂y}}{\frac{∂F}{∂z}}=-{\frac{3y^2}{-1}}=3y^2$

$\frac{∂z}{∂y} $ in $ P(1,1,2)=3$

$Dz(1,1,2)=(\frac{∂z}{∂x}(1,1,2) \qquad\frac{∂z}{∂y} (1,1,2)) $

$Dz(1,1,2)=(2 \qquad 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:

$D_{\underline x} f({\underline x°}) = -[D_{\underline y}F(\underline x°,\underline y°)]^{-1}D_\underline xF(\underline x°,\underline y°)$.

I don’t understand how this formula can be used in order to calculate $Dz$.

Any help is appreciated.

Thanks in advance.

Answer 1

It seems that both approaches are indeed equivalent.

Since $Dz$ is requested,

$D_{\underline x} f({\underline x°}) = -[D_{\underline y}F(\underline x°,\underline y°)]^{-1}D_\underline xF(\underline x°,\underline y°)$ becomes

$D_{\underline x, \underline y}f({\underline x°},{\underline y°}) = -[D_{\underline z}F(\underline x°,\underline y°\underline z°)]^{-1}D_{\underline x \underline y} F(\underline x°,\underline y°,\underline z°)$.

It follows that:

$ -[D_{\underline z}F(\underline x°,\underline y°\underline z°)]^{-1}=-[-1]^{-1}$

$D_{\underline x \underline y} F(\underline x°,\underline y°,\underline z°)= (2x\qquad 3x^2) $.

$D_{\underline x \underline y} F(1, 1, 2)= (2\qquad 3) $.

So,

$Dz = -[-1]^{-1}(2\qquad 3)=(2\qquad 3)$.