Changes

We have just met with the fact that the differential of the <math>\operatorname{and}</math> is the <math>\operatorname{or}</math> of the differentials.  Briefly summarized:

We have just met with the fact that the differential of the <math>\operatorname{and}</math> is the <math>\operatorname{or}</math> of the differentials.  Briefly summarized:

: <math>x\ \operatorname{and}\ y\ \xrightarrow{\operatorname{Diff}}\ \operatorname{d}x \ \operatorname{or}\ \operatorname{d}y</math>

: <p><math>x\ \operatorname{and}\ y\ \xrightarrow{\operatorname{Diff}}\ \operatorname{d}x \ \operatorname{or}\ \operatorname{d}y</math></p>

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Given the proposition ''f''(''x'', ''y'') in ''U'' = ''X''&nbsp;&times;&nbsp;''Y'', the (''first order'') ''difference'' of ''f'' is the proposition ''Df'' in ''EU'' that is defined by the formula ''Df'' = ''Ef''&nbsp;&ndash;&nbsp;''f'', or, written out in full, ''Df''(''x'', ''y'', ''dx'', ''dy'') = ''f''(''x'' + ''dx'', ''y'' + ''dy'')&nbsp;&ndash;&nbsp;''f''(''x'', ''y'').

Given the proposition <math>f(x, y)\!</math> in <math>U = X \times Y,</math> the (''first order'') ''difference'' of <math>f\!</math> is the proposition <math>\operatorname{D}f</math> in <math>\operatorname{E}U</math> that is defined by the formula <math>\operatorname{D}f = \operatorname{E}f - f,</math> or, written out in full, <math>\operatorname{D}f(x, y, \operatorname{d}x, \operatorname{d}y) = f(x + \operatorname{d}x, y + \operatorname{d}y) - f(x, y).</math>

In the example ''f''(''x'', ''y'') = ''xy'', the result is:

In the example <math>f(x, y) = xy,\!</math> the result is:

* ''Df''(''x'', ''y'', ''dx'', ''dy'') = (''x'' + ''dx'')(''y'' + ''dy'')&nbsp;&ndash;&nbsp;''xy''.

: <p><math>\operatorname{D}f(x, y, \operatorname{d}x, \operatorname{d}y) = (x + \operatorname{d}x)(y + \operatorname{d}y) - xy.</math></p>

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