I am working through implicit function theorem for the first time, and I have the following understanding. Given a system of $n$ equations, \begin{equation} f_i(x_1,\dots ,x_m,y_1,\dots , y_n)=0,\ \ \ \ \ \ \ i=1,\dots ,n \end{equation} With some point $p_0$ with coordinates $(a_1,\dots ,a_m,b_1,\dots ,b_n)$, and the condition, \begin{equation} \frac{\partial (f_1,\dots,f_n)}{\partial (y_1,\dots ,y_n)}\bigg|_{p_{0}}\neq 0 \end{equation} Then the system of equations can be solved for $y_1,\dots ,y_n$ as functions of $x_1,\dots ,x_m$ in the neighbourhood of the point. In this case the following equations hold, \begin{equation} f_i(x_1,\dots ,x_m,y_1(x_1,\dots ,x_m),\dots ,y_n(x_1,\dots ,x_m))=0,\ \ \ \ \ i=1,\dots ,n \end{equation} However I do not understand the following notation, which is the conclusion of the theorem! \begin{equation} \frac{\partial y_i}{\partial x_j}\bigg|_{i\neq j}=-\frac{\frac{\partial(f_1,f_2,\dots ,f_n)}{\partial (y_1,\dots ,x_j,\dots ,y_n)}}{\frac{\partial (f_1,f_2,\dots ,f_n)}{\partial (y_1,\dots ,y_i,\dots ,y_n)}} \end{equation} Could you please explain what these Jacobian determinants are, and why the $x_j$ and $y_i$ are in the denominators of each expression? Many thanks!

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    $\begingroup$ Firstly you say you don't understand the notation. Then you ask why the equality is true. What do you want to ask exactly? $\endgroup$
    – Git Gud
    Jul 7, 2015 at 23:15
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    $\begingroup$ Do you mean $i\neq j$ instead of $x\neq j$ on the lefthand side in the last equation? $\endgroup$
    – Arthur
    Jul 7, 2015 at 23:15
  • $\begingroup$ I understand implicit functions for only 2 or 3 variables so i do not ask why equity holds (apologies) but what the notation means. I think so, I will change that! thank you $\endgroup$
    – RedPen
    Jul 7, 2015 at 23:17
  • $\begingroup$ In short, to prove the theorem, you linearize the level mapping then focus on the particular submatrix to eliminate. Invertibility of the submatrix is captured by the Jacobian determinants. Don't be bothered by this, it's not magic, it's just the usual linear algebra and you can see the determinants as stemming from Cramer's Rule. I don't have that in my answer, but I hope there is enough to "see" it... $\endgroup$ Jul 8, 2015 at 6:48

1 Answer 1


I happen to have some notes on this, perhaps they help:

Given $n$-equations in $(m+n)$-unknowns when can we solve for the last $n$-variables as functions of the first $m$-variables ? Given a continuously differentiable mapping $G=(G_1,G_2,\dots , G_n): \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ study the level set: (here $k_1,k_2,\dots , k_n$ are constants) \begin{align} \notag G_1(x_1, \dots , x_m, y_1, \dots , y_n)&=k_1 \\ \notag G_2(x_1, \dots , x_m, y_1, \dots , y_n)&=k_2 \\ \notag & \vdots \\ \notag G_n(x_1, \dots , x_m, y_1, \dots , y_n)&=k_n \notag \end{align} We wish to locally solve for $y_1, \dots , y_n$ as functions of $x_1, \dots x_m$. That is, find a mapping $h : \mathbb{R}^m \rightarrow \mathbb{R}^n$ such that $G(x,y)=k$ iff $y=h(x)$ near some point $(a,b) \in \mathbb{R}^m \times \mathbb{R}^n$ such that $G(a,b)=k$. In this section we use the notation $x=(x_1,x_2,\dots x_m)$ and $y=(y_1,y_2,\dots , y_n)$.

Before we turn to the general problem let's analyze the unit-circle problem in this notation. We are given $G(x,y)=x^2+y^2$ and we wish to find $f(x)$ such that $y=f(x)$ solves $G(x,y)=1$. Differentiate with respect to $x$ and use the chain-rule: $$ \frac{\partial G}{\partial x}\frac{dx}{dx} + \frac{\partial G}{\partial y}\frac{dy}{dx} = 0 $$ We find that $\boxed{dy/dx = -G_x/G_y} = -x/y$. Given this analysis we should suspect that if we are given some level curve $G(x,y)=k$ then we may be able to solve for $y$ as a function of $x$ near $p$ if $G(p)=k$ and $G_y(p) \neq 0$. This suspicion is valid and it is one of the many consequences of the implicit function theorem.

We again turn to the linearization approximation. Suppose $G(x,y)=k$ where $x \in \mathbb{R}^m$ and $y \in \mathbb{R}^n$ and suppose $G: \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable. Suppose $(a,b) \in \mathbb{R}^m \times \mathbb{R}^n$ has $G(a,b)=k$. Replace $G$ with its linearization based at $(a,b)$: $$ G(x,y) \approx k + G'(a,b)(x-a,y-b) $$ here we have the matrix multiplication of the $n \times (m+n)$ matrix $G'(a,b)$ with the $(m+n) \times 1$ column vector $(x-a,y-b)$ to yield an $n$-component column vector. It is convenient to define partial derivatives with respect to a whole vector of variables, $$ \frac{\partial G}{\partial x} = \left[ \begin{array}{ccc} \tfrac{\partial G_1}{\partial x_1} & \cdots & \tfrac{\partial G_1}{\partial x_m} \\ \vdots & & \vdots \\ \tfrac{\partial G_n}{\partial x_1} & \cdots & \tfrac{\partial G_n}{\partial x_m} \end{array} \right] \qquad \frac{\partial G}{\partial y} = \left[ \begin{array}{ccc} \tfrac{\partial G_1}{\partial y_1} & \cdots & \tfrac{\partial G_1}{\partial y_n} \\ \vdots & & \vdots \\ \tfrac{\partial G_n}{\partial y_1} & \cdots & \tfrac{\partial G_n}{\partial y_n} \end{array} \right] $$ In this notation we can write the $n \times (m+n)$ matrix $G'(a,b)$ as the concatenation of the $n \times m$ matrix $\frac{\partial G}{\partial x}(a,b) $ and the $n \times n$ matrix $\frac{\partial G}{\partial y}(a,b)$ $$ G'(a,b) = \biggl[\frac{\partial G}{\partial x}(a,b) \bigg{|} \frac{\partial G}{\partial y}(a,b) \biggl] $$ Therefore, for points close to $(a,b)$ we have: $$ G(x,y) \approx k + \frac{\partial G}{\partial x}(a,b)(x-a)+\frac{\partial G}{\partial y}(a,b)(y-b) $$ The nonlinear problem $G(x,y)=k$ has been (locally) replaced by the linear problem of solving what follows for $y$: $$ k \approx k + \frac{\partial G}{\partial x}(a,b)(x-a)+\frac{\partial G}{\partial y}(a,b)(y-b) $$ Suppose the square matrix $\frac{\partial G}{\partial y}(a,b)$ is invertible at $(a,b)$ then we find the following approximation for the implicit solution of $G(x,y)=k$ for $y$ as a function of $x$: $$ y = b - \biggl[\frac{\partial G}{\partial y}(a,b) \biggr]^{-1}\biggl[\frac{\partial G}{\partial x}(a,b)(x-a) \biggl]. $$ Of course this is not a formal proof, but it does suggest that $det\bigl[\frac{\partial G}{\partial y}(a,b) \bigr] \neq 0$ is a necessary condition for solving for the $y$ variables.

As before suppose $G: \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^n$. Suppose we have a continuously differentiable function $h: \mathbb{R}^m \rightarrow \mathbb{R}^n$ such that $h(a)=b$ and $G(x,h(x))=k$. We seek to find the derivative of $h$ in terms of the derivative of $G$. This is a generalization of the implicit differentiation calculation we perform in calculus I. I'm including this to help you understand the notation a bit more before I state the implicit function theorem. Differentiate with respect to $x_l$ for $l \in \{1,2,\dots n\}$: $$ \frac{\partial}{\partial x_{l}} \biggl[ G(x,h(x)) \biggr] = \sum_{i=1}^{m}\frac{\partial G}{\partial x_i } \frac{\partial x_i}{\partial x_l } + \sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_l} = \frac{\partial G}{\partial x_l } + \sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_l} = 0$$ we made use of the identity $\frac{\partial x_i}{\partial x_k } = \delta_{ik}$ to squash the sum of $i$ to the single nontrivial term and the zero on the r.h.s follows from the fact that $\frac{\partial}{\partial x_l} (k)=0$. Concatenate these derivatives from $k=1$ up to $k=m$: $$ \biggl[ \frac{\partial G}{\partial x_1 } + \sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_1} \bigg{|} \frac{\partial G}{\partial x_2 } + \sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_2} \bigg{|} \cdots \bigg{|} \frac{\partial G}{\partial x_m } + \sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_m} \biggr] = [0|0| \cdots |0] $$ Properties of matrix addition allow us to parse the expression above as follows: $$ \biggl[ \frac{\partial G}{\partial x_1 } \bigg{|} \frac{\partial G}{\partial x_2 } \bigg{|} \cdots \bigg{|} \frac{\partial G}{\partial x_m } \biggr] + \biggl[ \sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_1} \bigg{|} \sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_2} \bigg{|} \cdots \bigg{|} \sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_m} \biggr] = [0|0| \cdots |0] $$ But, this reduces to $$ \frac{\partial G}{\partial x } + \biggl[ \frac{\partial G}{\partial y}\frac{\partial h}{\partial x_1} \bigg{|} \frac{\partial G}{\partial y}\frac{\partial h}{\partial x_2} \bigg{|} \cdots \bigg{|} \frac{\partial G}{\partial y}\frac{\partial h}{\partial x_m} \biggr] = 0 \in \mathbb{R}^{m \times n} $$ The concatenation property of matrix multiplication states $[Ab_1|Ab_2| \cdots | Ab_m] = A[b_1|b_2| \cdots | b_m]$ we use this to write the expression once more, $$ \frac{\partial G}{\partial x } + \frac{\partial G}{\partial y} \biggl[ \frac{\partial h}{\partial x_1} \bigg{|} \frac{\partial h}{\partial x_2} \bigg{|} \cdots \bigg{|} \frac{\partial h}{\partial x_m} \biggr] = 0 \ \ \Rightarrow \ \ \frac{\partial G}{\partial x } + \frac{\partial G}{\partial y} \frac{\partial h}{\partial x} = 0 \ \ \Rightarrow \ \ \boxed{\frac{\partial h}{\partial x} = -\frac{\partial G}{\partial y}^{-1}\frac{\partial G}{\partial x }} $$ where in the last implication we made use of the assumption that $\frac{\partial G}{\partial y}$ is invertible.


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