Implicit function theorem, product of partial derivaticves

[LastMinute]

Homework Statement

f(x) = f(x_1,...,x_N) : R^N \mapsto R has continuous partial derivatives. Assume that for a point a in R^N , \frac{\partial f}{\partial x_i}(a) \neq 0 for all i = 1...N.

The implicit function theorem says that near a, the equation f(x) = f(a) can be used to express each x_i in terms of other x_j s. Prove the following equality at a:

Homework Equations

(-1)^N = \frac{\partial x_1}{\partial x_2} \cdot \frac{\partial x_2}{\partial x_3} ... \cdot \frac{\partial x_N-1}{\partial x_N} \cdot \frac{\partial x_N}{\partial x_1}

\frac{\partial}{\partial y_j} f(g(y)) = \nabla f(g(y)) \cdot \frac{\partial}{\partial y_j} g(y) (chain rule)

The Attempt at a Solution

I've tried writing something like x_1 = x_1(x_2,...,x_N) to make each one a function of the other and tried using the chain rule on that, but it gets really messy quickly. I'm really unsure about the significance of the implicit function theorem or where to start with this problem. I found a similar law called the triple product rule which seems to be this equality in N = 3, but I've been unable to extend it to general N. Any help would be greatly appreciated.

 

[Fredrik]

The implicit function theorem is used when you're not given a function, but are instead given a relationship between variables that you can use to define a function.

The statement that the implicit function theorem says that f(x)=f(a) implies a relationship between the $x_i$ seems wrong to me. Maybe you left out some relevant information? For some functions f, all you will get from f(x)=f(a) is x=a.

 

[LastMinute]

Yeah, I didn't quite get that either. I've typed out the question exactly as it appears. Is there a way to do it if maybe we just ignore what they've said about f(x) = f(a) and try to prove the result another way?

 

[mentats]

I think we have all the information we need to prove it. It seems to me that the solution to this lies in applying:

\frac{\partial x_j}{\partial x_k}=-\frac{\partial f/ \partial x_k}{\partial f/ \partial x_j}

However, I haven't been able to find a proof that involves the general n-variable case, even though many websites take it as a given. I have tried working it out on my own by setting z=f(x), taking \frac{\partial z}{\partial x_j}, applying the chain rule and rearranging so that the LHS is the same as above, but I am struggling to justify how all the partials that don't involve x_j and x_k are zero when we set z=f(a) (ie keep z constant).

 

[Fredrik]

I thought about this some more and I think I finally see what's going on in this problem. Since f is differentiable at a, f is continuous at a. So the fact that all the partial derivatives at a are non-zero, implies that there's an open set U that's a subset of the domain of f, contains a, and is such that the partial derivatives of f are all non-zero at every point in U. For every t in f(U), the equality $t=f(x_1,\dots,x_N)$ defines a relationship between the N variables $x_1,\dots,x_N$. Now the implicit function theorem guarantees the existence of functions $g_1,\cdots,g_N$ such that $x_1=g_1(x_2,\dots,x_N)$ and so on.

For example, consider f(x,y,z)=x+y+2z. (This function is so simple that we don't even need the implicit function theorem). Its partial derivatives are all non-zero at (0,0,0). We can take $U=\mathbb R^3$. Now let $t\in f(U)$ be arbitrary. The equality f(x,y,z)=t implies that x=t-y-2z, which implies that $\partial x/\partial y=-1$. The other partial derivatives can be computed similarly. So
$$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=(-1)(-2)\left(-\frac{1}{2}\right) =-1.$$ This is of course only a special case. What you need to do to solve the problem is to figure out what the implicit function theorem says about the partial derivatives of the functions defined implicitly by $f(x_1,\dots,x_N)=t$ where b is some point in f(U).

Note that the times when we really need the implicit function theorem are when we can't solve for each variable the way I did in the example.