I apologize in advance as this is rather roundabout. I have seen the Implicit Function Theorem (in a more specific case) stated as such

Let $x=(x_1, x_2, \ldots, x_m)$. Let $G(x, y): \mathbb{R}^{m+1} \to \mathbb{R}$ be a once continuously differentiable function around some neighborhood $N_\delta$ of $(x_0, y_0)$. If $G(x_0, y_0) = c$ and $\dfrac{\partial G(x, y)}{\partial y}|_{(x_0, y_0)} \neq 0$ then there exists a once differentiable $y(x)$ on some neighborhood $N_\varepsilon$ of $x_0$ such that

  1. $G(x, y(x)) = c \; \forall x \in N_\varepsilon$
  2. $y(x_0) = y_0$
  3. $\dfrac{\partial y}{\partial x_i} |_{x_0} = -\dfrac{\dfrac{\partial G}{\partial x_i}|_{(x_0, y_0)}}{\dfrac{\partial G}{\partial y}|_{(x_0, y_0)}}$

I am hesitant accepting the third antecedent of the theorem. Assuming 1,2 are true. We have by chain rule \begin{align} 0 &= \sum_{j=1}^{m} \dfrac{\partial G}{\partial x_j} \dfrac{d x_j}{d x_i} + \dfrac{\partial G}{\partial y} \dfrac{d y}{d x_i}\\ \dfrac{d y}{d x_i} &= -\dfrac{\sum_{j=1}^{m} \dfrac{\partial G}{\partial x_j} \dfrac{d x_j}{d x_i}}{\dfrac{\partial G}{\partial y}} \end{align}

This seems to be a much weaker result. The question is: Why does the left hand side of (3) have partial derivatives and not total derivatives? Further, why does it seem to be $\dfrac{d x_j}{d x_i} = 0$ for $i \neq j$ so that most of the terms of the sum are removed? I suspect it may have to do with wrongly manipulating the chain rule. Most proofs of the theorem that have been shown to me lack the detail and simply conclude (3) "by implicit differentiation."

  • 1
    $\begingroup$ In the implicit function theorem, the $x_i$'s are independent coordinates, so they don't depend on each other. This is why $\frac{dx_j}{dx_i}=0$ for $i\not=j$. $\endgroup$ Oct 9, 2018 at 0:06
  • $\begingroup$ @MichaelBurr Thanks, that was my suspicion. That seems a bit subtle to me. Is there anything in the theorem that should lead me to that assumption (or perhaps my statement of the theorem is incomplete)? $\endgroup$
    – Almacomet
    Oct 9, 2018 at 0:10

1 Answer 1


In the implicit function theorem, you are looking at $G(x,y)$ where $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_m)$ and trying to determine if each $y_j$ is a function of the $x_i$'s. In other words, the implicit function theorem is trying to determine if one can write $y_j=y_j(x_1,\dots,x_n)$. In your case above, $y$ might consist of a single variable instead of a vector.

In this setup, the $x_i$'s are independent variables, the inputs to the function. Therefore, changing one of the $x_i$'s doesn't require a change in any of the other $x_i$'s. Therefore, $$ \frac{dx_j}{dx_i}=0 $$ for $i\not=j$. $x_j$ simply isn't a function of $x_i$ (unless $i=j$).

In your right-hand-side, I would not write $\frac{dy}{dx_i}$, the reason is that $y$ is meant to be a function of several of the $x_i$'s, so the derivative should be a partial derivative, and not a regular derivative. The chain rule doesn't always end with a regular derivative, it depends on the dependence (and number) of the variables.


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