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**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

**R**x

**R**? What about the domains of the exact derivative function of

*f*,

[tex]\frac{\mathrm{d} f}{\mathrm{d} x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial g} \frac{\mathrm{d} g}{\mathrm{d} x},[/tex]

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,

[tex]g(x)=x^2;[/tex]

[tex]f(x,g(x)) = x \cdot g(x) = x \cdot x^2 = x^3.[/tex]

Then, differentiating first wrt

*g*, and afterwards substituting

*x*

^{2}for

*g(x)*,

[tex]\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)[/tex]

[tex]=g(x)+x \cdot 2x = x^2 + 2x^2 = 3x^2,[/tex]

gives the same result as substituting first, only if the embedded function

*g*is "opaque" to the partial operator:

[tex]\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.[/tex]

But is it opaque? If we suppose the partial derivative can "see through"

*g(x)*, allowing us to substitute

*x*

^{2}for

*g(x)*everywhere, then the result differs. If I've got this right, where

*x*is positive, for example:

[tex]\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)[/tex]

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

where, for

*x*positive,

[tex]\frac{\partial (x^3)}{\partial (x^2)} = \frac{\partial (x^{3/2})}{\partial x}.[/tex]

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