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Let $x = (x_{1}, x_{2}, x_{3})^{T} \in \mathbb{R}^{3}$, and take a sufficiently differentiable function $f: \mathbb{R}^{3} \mapsto \mathbb{R}$.

Suppose that I am searching for zeroes of $f$ such that I can express $x_{3}$ in terms of $x_{1}$ and $x_{2}$. In other words I am looking at the level set $f=0$ and by implicit function theorem I require that $\frac{\partial f}{\partial x_{3}} \neq 0$ on that level set.

Suppose that $f$ is such that I can $\textbf{always}$ express $x_{1}$ explicitly in terms of other two variables, i.e. can write $x_{1} = g(x_{2}, x_{3})$ for some $g$.

Is there a way to express the condition $\frac{\partial f}{\partial x_{3}} \neq 0$ in terms of $g$?

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Lets be careful of the variables, so you can keep track of what is what. On $\Bbb R^2$, use $(y_1, y_2)$ and on $\Bbb R^3$, use $(x_1, x_2, x_3)$ Now consider the map $$h : \Bbb R^2 \to \Bbb R^3 : (y_1, y_2) \mapsto (x_1, x_2, x_3) := (g(y_1, y_2), y_1, y_2)$$

Then $$\begin{array}{cc}\frac{\partial x_1}{\partial y_1} = \frac{\partial g}{\partial y_1} & \frac{\partial x_1}{\partial y_2} = \frac{\partial g}{\partial y_2}\\\frac{\partial x_3}{\partial y_1} = 1 & \frac{\partial x_1}{\partial y_2} = 0\\\frac{\partial x_2}{\partial y_1} = 0 & \frac{\partial x_1}{\partial y_2} = 1\end{array}$$

Now $f\circ h \equiv 0$, so $$0 = \frac{\partial f\circ h}{\partial y_2} = \left.\frac{\partial f}{\partial x_1}\right|_h\frac{\partial x_1}{\partial y_2} + \left.\frac{\partial f}{\partial x_2}\right|_h\frac{\partial x_2}{\partial y_2} + \left.\frac{\partial f}{\partial x_3}\right|_h\frac{\partial x_3}{\partial y_2}\\=\left.\frac{\partial f}{\partial x_1}\right|_h\frac{\partial g}{\partial y_2} + \left.\frac{\partial f}{\partial x_3}\right|_h$$ So $\left.\frac{\partial f}{\partial x_3}\right|_h = 0$ if and only if $\left.\frac{\partial f}{\partial x_1}\right|_h\frac{\partial g}{\partial y_2} = 0$ if and only if $\left(\left.\frac{\partial f}{\partial x_1}\right|_h = 0\right.$ or $\left.\frac{\partial g}{\partial y_2} = 0\right)$.

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  • $\begingroup$ Thanks, that is very useful and insightful. $\endgroup$ – Alex May 17 '16 at 12:30

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