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::<math>f' = f'.\,</math>
:It can be said that just one factor is a derivative, and it is [[vacuous truth|vacuously true]] that the others are not, because there are
*To go from case ''n'' to case ''n'' + 1 is trivial if ''n'' ≥ 2, but impossible if ''n'' = 1.
*To go from case ''n'' to case ''n'' + 1, one must ''use'' the case ''n'' = 2 (i.e. one must use the ordinary product rule taught in differential calculus).
*The substantial part of the proof is the case ''n'' = 2, and that is precisely the product rule that is taught differential calculus.
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The usual [[triangle inequality]] in [[metric space]]s says
:<math>d(
For a sequence of ''n'' points, one has
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:<math>d(x_1,x_n) \le d(x_1,x_2)+\cdots+d(x_{n-1},x_n).\,</math>
Now observe four facts
* The ''first'' case is
::<math>d(x_1,x_2) \le d(x_1,x_2),\,</math>
: and this is vacuously true, in that it does not rely on any information about what kind of function ''d'' is (i.e., that ''d'' is a metric).
* The ''n''th case is the case with ''n'' terms on the right side, and that is
::d(x_1,x_{n+1}) \le d(x_1,x_2) + \cdots + d(x_n, x_{n+1}).\,
: To go from there to the (''n'' + 1)th case is trivial if ''n'' ≥ 2, but impossible if ''n'' = 1.
* To go from case ''n'' to case ''n'' + 1, one must ''use'' the case ''n'' = 2 (i.e. one must use the ordinary triangle inequality).
* The substantial part of the proof is the ''second'' case. The second case is just the statement of the ordinary triangle inequality, and is different in different metric spaces. It is part of the proof that a particular function ''d'' is in fact a metric. In some cases is difficult or otherwise onerous. Usually the other aspects of proving that ''d'' is a metric are trivial (i.e. that ''d''(''x'', ''x'') = 0 for all ''x'', and that ''d'' is symmetric).
[More examples to be added.]
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