Derivative when you just have Constraints

[mXSCNT]

Suppose that you have a set of real variables {x₁,x₂,...,x_n}. If x₁ = f(x₂,x₃,...,x_n) then this represents a constraint on all the variables. In this case, it's possible to find dx₁/dx_i as long as f is differentiable. But not all possible constraints among the x_j are of this form. How might one extend the concept of derivative to the case when the constraint on all the variables is not necessarily that one variable is a function of the others?

 

[yyat]

Welcome to the world of http://en.wikipedia.org/wiki/Differential_manifold" !

Suppose you have a differentiable function $$g:\mathbb{R}^n\to\mathbb{R}^k$$ and your constraints are given by the equation g(x)=0 (this is the general form). If the function g is "nice", for example if 0 is a regular value of g, then the set where g(x)=0 is a differentiable manifold of dimension n-k. As an example, take $$g(x)=x_1^2+x_2^2-1$$, then the resulting manifold is the unit circle.
One can then define the derivative of a differentiable function between manifolds abstractly, but to get partial derivatives (and do concrete calculations) you need to choose local coordinates (aka a chart) on the manifold.

 

[mXSCNT]

Does your example allow you to say that $$\frac{dx_1}{dx_2}\big|_{x_1=x_2=\frac{1}{\sqrt(2)}} = -1$$?

 

[yyat]

Yes, $$x_1$$ defines a real-valued function on the unit circle, and near the point $$(1/\sqrt{2},1/\sqrt{2})$$ one can "parametrize" the circle by $$x_2$$, i.e. $$x_2$$ gives differentiable local coordinates. In these coordinates $$x_1=\sqrt{1-x_2^2}$$ and you can simply compute the usual derivative.