# Sufficient conditions using implicit function theorem

Let $$f:\mathbb{R^3} \to \mathbb{R}.$$

Let $$(x_0,y_0,z_0)$$ be a solution of the equation $$f(x-y,y-z,z-x)=0$$.

Find sufficient conditions such that extracting $$z$$ with respect to $$x,y$$ is possible.

I study about implicit function theorem and I find out this theorem but the equation $$f(x-y,y-z,z-x)=0$$ gets me a little bit confused.

My attempt:

For applying implicit function theorem in $$(a,b,c)$$ I have to demand:

1.$$f(a,b,c)=0$$.

2.There is a neigborhood $$U$$ of $$(a,b,c)$$ such that $$f$$ is $$C^1$$ in $$U$$ and

3.$$\frac{\partial f}{\partial z}(a,b,c)\neq 0$$

Denote $$a:=x_0-y_0,b:=y_0-z_0,c:=z_0-x_0 \implies f(a,b,c)=0$$.

$$f\in C^1$$.

I have an issue finding the derivative of $$f$$.

Is my solution correct?

• It is not $\partial f/\partial z$ which must be non-zero at $(a,b,c).$ It is $\partial g/\partial z,$ where $g(x,y,z)=f(x-y,y-z,z-x).$ To compute $\partial g/\partial z,$ use the chain rule: $\partial g/\partial z(x,y,z)=(\partial_3f-\partial_2f)((x-y,y-z,z-x).$ Feb 11, 2023 at 15:53
• @AnneBauval Sorry , you are right , fixed
– Algo
Feb 11, 2023 at 15:55

You can define $$F(x,y,z) = f(x-y,y-z,z-x)$$ and $$\phi(x,y,z)= (x-y, y-z,z-x)$$. You have $$F(x,y,z) = (f \circ \phi)(x,y,z)$$. From there, you just have to apply the chain rule:
$$\frac{\partial F}{\partial z}(x,y,z) = -\frac{\partial f}{\partial y}(x-y,y-z,z-x) + \frac{\partial f}{\partial z}(x-y,y-z,z-x).$$ Finally, the condition you're looking for is
$$\frac{\partial f}{\partial y}(x_0-y_0,y_0-z_0,z_0-x_0) \neq \frac{\partial f}{\partial z}(x_0-y_,y_0-z_0,z_0-x_0)$$ applying implicit function theorem.
• thanks , did you take derivative of $y$ since there is $y-z$ is the second coordinate ? I meant , in case $y$ doesn't appear in the second coordinate of $f$, let's say its only $z$
It is not $$\frac{\partial f}{\partial z}(a,b,c)$$ which must be non-zero. It is $$\frac{\partial g}{\partial z}(x_0,y_0,z_0),$$ where $$g(x,y,z)=f(x-y,y-z,z-x).$$ To compute $$\frac{\partial g}{\partial z},$$ use the chain rule: $$\frac{\partial g}{\partial z}(x,y,z)=(\partial_3f-\partial_2f)(x-y,y-z,z-x).$$