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

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

Watching http://www.khanacademy.org/video/exact-equations-intuition-1--proofy?playlist=Differential%20Equations [Broken] 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,

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

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arildno
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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).

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

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.