Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a $C^1$ function,and $a\in \mathbb{R}^n$ such that $\frac{\partial f}{\partial x_i}(a)\ne 0$ for all $i$ and $f(a)=0$.

a) Prove that there is a neighbour of $a$ such that the equation $f=0$ fix $n$ functions:


b) Find $\Pi^n_{i=1}\frac{\partial x_i}{\partial x_{i+1}}$, where the indices are $mod(n)$ and all of the relevant derivatives are at $a$.

I have absolutely no idea how to begin...any ideas?

  • $\begingroup$ What is part a asking? $\endgroup$
    – Red shoes
    Jul 3, 2017 at 19:40
  • $\begingroup$ @Ashkan Edited. $\endgroup$
    – ChikChak
    Jul 3, 2017 at 19:49
  • $\begingroup$ The last term in the product at b involves $x_{n+1}$. Are we supposed to assume $x_{n+1}=x_1$ (that's my guess)? $\endgroup$ Jul 3, 2017 at 19:58
  • $\begingroup$ @HaraldHanche-Olsen yes, bcz It said $mod(n)$ $\endgroup$
    – Red shoes
    Jul 3, 2017 at 20:02
  • 1
    $\begingroup$ For your second question, take a look at this. $\endgroup$ Jul 3, 2017 at 21:29

2 Answers 2


Il order not to mix everything let us denote $X_i$ the new functions, more precisely as $\partial f\over \partial x_i$$ \not =0$, there exists near $(a_1,....a_{i-1},a_{i+1},...a_n)$ a function $X_i (x_1,...,x_{i-1},x_{i+1},...x_n)$ such that $f(x_1,...,x_{i-1},X_i, x_{i+1},...x_n)=0$ and $X_i (a_1,...,a_{i-1},a_{i+1},...a_n)=a_i$.

Deriving the equation $X_i (x_1,...,x_{i-1},x_{i+1},...x_n)$ by $x_{i+1}$ we get :

${\partial f \over \partial x_i} (a).{\partial X_{i+1} \over \partial x_i} +{\partial f \over \partial x_{i+1}} (a)=0 .$

Then ${\partial X_{i+1} \over \partial x_i}= -{ {\partial f \over \partial x_{i+1}} (a)\over {\partial f \over \partial x_{i}} (a)}$

We deduce the product :

$\Pi_1^n{\partial X_{i+1} \over \partial x_i}= (-1)^n$



For part one, Take $i$ and Apply implicit function theorem on equation $f(x_1 x_2,...,x_n)=0$ at point $x=a$ to represent $x_i$ as function of $(x_1,..,x_{i-1}, x_{i+1}..,x_n)$ i.e., $x_i(x_1,..,x_{i-1}, x_{i+1}..,x_n)$. Note that you know $\frac{\partial f}{\partial x_i}(a)\ne 0$.

For part b) use chain rule .

Note that $\Pi^n_{i=1}\frac{\partial x_i}{\partial x_{i+1}} =\frac{\partial x_1}{\partial x_{1}} =1 $

  • $\begingroup$ You're totally wrong on part b. You need to think through implicit differentiation. $\endgroup$ Jul 3, 2017 at 21:20
  • $\begingroup$ @TedShifrin you're saying final answer is not 1 in part b. In both parts I didn't write complete solution , I just wanted give OP hint how to start problem! $\endgroup$
    – Red shoes
    Jul 4, 2017 at 3:44
  • $\begingroup$ You don't know the correct answer (or proof) for part b. Look at the questioin I linked in the comments to the OP. $\endgroup$ Jul 4, 2017 at 5:17
  • $\begingroup$ @TedShifrin Can we say if $n$ is even number then the answer is $1$? $\endgroup$
    – Red shoes
    Jul 5, 2017 at 4:06

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