# Implicit Differentiation - Different Approaches 2

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.

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,

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