Differential $\mathrm{d}^2f$ of implicit function $F(x,y,z)=xyz-x-y-z=0$

Question

Determine the differential $\mathrm{d}^2f$ of the implicit function defined as $z=f(x,y)$:

$$F(x,y,z)=xyz-x-y-z=0$$

So in fact of the implicit function I have to use the implicit function theorem for $f'(x_0,y_0,z_0)$, I get: $$f'(x_0,y_0,z_0)=\frac{1}{xyz+x+y+z}\cdot \left(-xyz-x-y-z,-xyz-x-y-z\right)=(-1,-1)$$

In my script there is the definition of the differential: $h \in \mathbb{R}: d_hf(x_0,y_0,z_0=f'(x_0,y_0,z_0) \cdot h$.

Applied on the derivate of the implicit function I get: $$d_hf(x_0,y_0,z_0)=(-1,-1)\cdot h=2h$$

And here I'm taken aback. This can't be the right thing but I couldn't see a mistake in derivatives.

Any hints? Thanks for your help!

Answer 1

$$xyz-x-y-z=0$$

for the first order derivatives, $$yz+xy\frac{\partial f}{\partial x}-1-\frac{\partial f}{\partial x}=0$$ $$xz+xy\frac{\partial f}{\partial y}-1-\frac{\partial f}{\partial y}=0$$

This gives, $$\frac{\partial f}{\partial x}=\frac{1-yz}{xy-1}$$ $$\frac{\partial f}{\partial y}=\frac{1-xz}{xy-1}$$

for the second order derivatives, $$y\frac{\partial f}{\partial x}+xy\frac{\partial^2 f}{\partial x^2}-\frac{\partial^2 f}{\partial x^2}=0$$

$$x\frac{\partial f}{\partial y}+xy\frac{\partial^2 f}{\partial y^2}-\frac{\partial^2 f}{\partial y^2}=0$$

$$z+x\frac{\partial f}{\partial x}+xy\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial x\partial y}=0$$

This gives, $$\frac{\partial^2 f}{\partial x^2}=\frac{y\frac {\partial f}{\partial x}}{1-xy}$$ $$\frac{\partial^2 f}{\partial y^2}=\frac{x\frac {\partial f}{\partial y}}{1-xy}$$ $$\frac{\partial^2 f}{\partial x\partial y}=\frac{z+x\frac {\partial f}{\partial y}}{1-xy}$$

Answer 2

First of all, $\frac{\partial z}{\partial x} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} = -\frac{yz - 1}{xy - 1}$ and similarly $\frac{\partial z}{\partial y} = -\frac{xz - 1}{xy - 1}$.

So, $df = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy = \frac{1 - yz}{xy - 1} dx + \frac{1 - xz}{xy - 1} dy$.

See if you can compute $d^2 f$ from here.