View source for Doctrine:Differentiation notation & terminology

==Problem==
Almost everyone has a different view of how to write derivatives. I have compiled these as "the best of the best" and tweaked it a little to enhance readability and consistency of the notation.
==Notation & terminology==
===Derivative===
* {{M|\mathrm{d}f\vert_a}} - the derivative of {{M|f}} at {{M|a}}. A [[linear map]] from the domain of {{M|f}} to the co-domain of {{M|f}}
** This means {{M|f}} is differentiable at {{M|a}} and {{M|\mathrm{d}f\vert_a}} is its derivative; '''''not''' differential'', that is something else.
We use the {{M|\mathrm{d} }} and the {{M|\vert}} as brackets. Everything between is what we're taking the derivative of. 
* '''Example: The chain rule''' - {{M|\mathrm{d}(g\circ f)\big\vert_{a}\eq\mathrm{d}g\big\vert_{f(a)}\circ\mathrm{d}f\big\vert_a}}<ref group="Note">We don't need the brackets around {{M|g\circ f}} however we can agree that this is easier to read than:
* {{M|\mathrm{d}g\circ f\big\vert_{a}\eq\mathrm{d}g\big\vert_{f(a)}\circ\mathrm{d}f\big\vert_a}}</ref>
** As opposed to: {{M|1=D(f\circ g)(a)=Dg(f(a))\circ Df(a)}}{{rAOMJRM}}{{rCOMS}}
The notation employed by Munkres and Spivak in {{M|Df(a)}} makes it hard to tell if it is referring to the derivative of {{M|f(a)}} considered as a function (which may be constant, or {{M|f}} might map {{M|a}} to a function that can be differentiated itself!) and if so where. This is especially true when looking at functions of functions.

{{Todo|This paragraph is repetitive, fix it}}
It is common to want to consider {{M|\mathrm{d}f\vert}}, which takes points in the domain of {{M|f}} to their derivatives at that point. This is a slight abuse of notation however in our notation it is not a big leap to see {{M|\mathrm{d}f\vert}} as a function that takes the domain of {{M|f}} to a function (from the domain of {{M|f}} to the co-domain of {{M|f}}). Then:
* {{M|\mathrm{d}f\vert(a)}} simply reads like another way of saying {{M|\mathrm{d}f\vert_a}}. 

==Overview==
Let {{M|(X,\Vert\cdot\Vert_X)}} and {{M|(Y,\Vert\cdot\Vert_Y)}} be [[normed spaces]] and let {{M|f:X\rightarrow Y}} be a [[function]]. Then:
* {{M|f}} is ''[[differentiable]]'' at {{M|a\in X}} if:
*# There exists a [[neighbourhood]] of {{M|a}}, {{M|N}} and
*# There exists a [[linear map]], {{M|\lambda:X\rightarrow Y}}<ref group="Note">Recall that {{M|X}} and {{M|Y}} are [[vector spaces]]</ref>
such that:
* {{MM|\lim_{h\rightarrow 0}\left(\frac{\Vert f(a+h)-f(a)-\lambda(h)\Vert_Y}{\Vert h\Vert_X}\right)\eq 0}} - {{Caution|There are other expressions for this limit that may be equivalent. This has not been decided/confirmed yet}}
{{Begin Notebox}}
'''Alternate expressions:'''
{{Begin Notebox Content}}
# {{MM|\lim_{h\rightarrow 0}\left(\frac{f(a+h)-f(a)-\lambda(h)}{\Vert h\Vert_X}\right)\eq\mathbf{0} }} - {{caution|unconfirmed}}
{{End Notebox Content}}{{End Notebox}}
==Caveats/Pending questions==
# Is this limit equivalent to the definition of differentiable that is least constraining?
# Are there normed subspaces? If so must it contain {{M|0}} (so {{M|h}} may tend to it?) - additionally if the neighbourhood of {{M|a}} is open in {{M|X}} and {{M|X}} is simply a [[topological subspace]] the requirements become blurred.
#* A normed subspace is almost certainly a vector subspace in its own right, so this shouldn't be a problem. However I mention this to remind myself, and as food for thought.
==Notes==
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==References==
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