Domain of f(x,g(g)), and partial derivatives

[Rasalhague]

Domain of f(x,g(x)), and partial derivatives

Watching http://www.khanacademy.org/video/exact-equations-intuition-1--proofy?playlist=Differential%20Equations Khan Academy video on exact equations, I got to wondering: if x is a real number, what is the domain of a function defined by f(x,g(x))? Is it R, or RxR? What about the domains of the exact derivative function of f,

$$\frac{\mathrm{d} f}{\mathrm{d} x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial g} \frac{\mathrm{d} g}{\mathrm{d} x},$$

and the partial derivative functions? It seems kind-of paradoxical to talk about the partial derivative of a function defined by f(x,g(x)); how can we hold x constant and vary g(x), or vice versa, without changing the definition of f(x,g(x)) to f(x,g(t)), where x and t are independent variables (or is that, in fact, what's done)?

Suppose, for example,

$$g(x)=x^2;$$

$$f(x,g(x)) = x \cdot g(x) = x \cdot x^2 = x^3.$$

Then, differentiating first wrt g, and afterwards substituting x² for g(x),

$$\frac{\partial }{\partial x} (x \cdot g(x)) + \frac{\partial }{\partial g}(x \cdot g(x)) \cdot \frac{\mathrm{d} g}{\mathrm{d} x}(x)$$

$$=g(x)+x \cdot 2x = x^2 + 2x^2 = 3x^2,$$

gives the same result as substituting first, only if the embedded function g is "opaque" to the partial operator:

$$\frac{\partial (x^3)}{\partial x}(x) + \frac{\partial (x^3)}{\partial g}(x) \cdot \frac{\mathrm{d} (x^2)}{\mathrm{d} x}(x) = 3x^2 + 0 \cdot 2x = 3x^2.$$

But is it opaque? If we suppose the partial derivative can "see through" g(x), allowing us to substitute x² for g(x) everywhere, then the result differs. If I've got this right, where x is positive, for example:

$$\frac{\partial (x^3)}{\partial x}(x) + \frac{\partial (x^{3/2})}{\partial x}(x) \cdot \frac{\mathrm{d} (x^2)}{\mathrm{d} x}(x)$$

$$= 3x^2 + \frac{3}{2}x^{1/2} \cdot 2x = 3(x^2+x^{3/2}),$$

where, for x positive,

$$\frac{\partial (x^3)}{\partial (x^2)} = \frac{\partial (x^{3/2})}{\partial x}.$$

 

[arildno]

The domain of h(x)=f(x,g(x)) is R. f is a two-variable function f(x,y), with domain R*R
The rate of change, dh/dx, is simply the rate of change of f experienced as we traverse the curve y=g(x).

 

[Rasalhague]

Thanks, arildno. I'm still a bit confused about what consequences this has for calculating partial derivatives. Are partial derivative functions undefined for h (at least, two distinct partial derivative functions)? When y=g(x), can we say that f = h:R-->R? Or does it need to still be regarded as f:RxR-->R for partial derivatives with respect to x and y to be meaningful?

 

[arildno]

Okay, we will be very precise about this.
This will mean that I choose to introduce new letters for variables that I didn't use in the previous post.

First, we have f(x,y) that is a mapping R*R on R
We also have a vector function, $$\vec{v}(t)=(v_{1}(t), v_{2}(t))=(t, g(t))$$
This is a mapping from R to R*R
Thus, we may define the composite function:
$$h(t)=f(\vec{v}(t)}$$
This is a mapping from R to R, going as follows R-->R*R-->R

Thus, we get:
$$\frac{dh}{dt}=\frac{\partial{f}}{\partial{x}}\frac{dv_{1}}{dt}+\frac{\partial{f}}{\partial{y}}\frac{dv_{2}}{dt}$$

To answer your questions:
1. "Are partial derivative functions undefined for h"
h being only a one-variable function do not have what we ordinarily calls "partial" derivatives. It only has a derivative with respect to its variable.

2. When y=g(x), can we say that f = h:R-->R?
h is R-->R, f is always R*R-->R
Even though x and y are interrelated by g, the arguments of f will always be two, namely "x" itself and g's function value.

Not that the domain of f(x,y) is the whole xy-plane. When we draw the graph (x,g(x)) there, then the function values of h(x) are simply f's function values computed along that graph.

 

[Rasalhague]

Thanks you, that's very helpful!

One more question, for now: which is the potential function, f or h?

(In the context of exact differential equations, that is. This article seems to use the same symbol, F, for both f and h.)