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Revision as of 03:32, 11 March 2009
, 03:32, 11 March 2009→Computation Summary : <math>f(u, v) = \texttt{((u)(v))}</math>: markup
| <math>\texttt{uv} \cdot \texttt{du~dv} ~+~ \texttt{u(v)} \cdot \texttt{du(dv)} ~+~ \texttt{(u)v} \cdot \texttt{(du)dv} ~+~ \texttt{(u)(v)} \cdot \texttt{((du)(dv))}</math>
| <math>\texttt{uv} \cdot \texttt{du~dv} ~+~ \texttt{u(v)} \cdot \texttt{du(dv)} ~+~ \texttt{(u)v} \cdot \texttt{(du)dv} ~+~ \texttt{(u)(v)} \cdot \texttt{((du)(dv))}</math>
|}
|}
<math>\operatorname{D}f</math> tells you what you would have to do, from where you are in the universe <math>[u, v],\!</math> if you want to bring about a change in the value of <math>f,\!</math> that is, if you want to get to a place where the value of <math>f\!</math> is different from what it is where you are. In the present case, where the reigning proposition <math>f\!</math> is <math>\texttt{((u)(v))},</math> the term <math>\texttt{uv} \cdot \texttt{du~dv}</math> of <math>\operatorname{D}f</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then you would have to change both <math>u\!</math> and <math>v\!</math> in order to reach a place where the value of <math>f\!</math> is different from what it is where you are.
<pre>
<pre>
Figure 1.4 approximates Df by the linear form
Figure 1.4 approximates Df by the linear form
df = uv 0 + u(v) du + (u)v dv + (u)(v)(du, dv).
df = uv 0 + u(v) du + (u)v dv + (u)(v)(du, dv).
