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The contents of the List of differentiation identities page were merged into Differentiation rules on February 6, 2011. For the contribution history and old versions of the redirected page, please see its history; for the discussion at that location, see its talk page.

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completed merge, some assistance required

I merged List of List of differentiation identities here except for this:

For higher derivatives the chain rule is given by Faà di Bruno's formula (below is the combinatoric form):

${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum _{\pi \in \Pi }f^{(\left|\pi \right|)}(g(x))\cdot \prod _{B\in \pi }g^{(\left|B\right|)}(x)}$

This belongs in the Nth derivatives section, but someone who knows more about math needs to do this. D O N D E groovily Talk to me 17:39, 6 February 2011 (UTC)

Possible error ?

shouldn't the n'th derivative of the power x^N be proportional to x^(N-n) instead of the product index r ? —Preceding unsigned comment added by 132.187.40.95 (talk) 11:20, 11 April 2011 (UTC)

Thanks very much for finding that ! ... Maschen (talk) 12:20, 8 June 2011 (UTC)

The polynomial or elementary power rule

Right now this section states: "If ${\displaystyle f(x)=x^{n}}$, for some natural number n (including zero) then

${\displaystyle f'(x)=nx^{n-1}.\,}$"

But this isn't only true for ${\displaystyle n\in \mathbb {N} }$; it's true for any ${\displaystyle n\in \mathbb {C} }$ (and of course ${\displaystyle \mathbb {R} }$, ${\displaystyle \mathbb {N} }$, ${\displaystyle \mathbb {Q} }$, ${\displaystyle \mathbb {Z} }$ etc.). — Preceding unsigned comment added by Philmac88 (talk • contribs) 20:54, 16 August 2011 (UTC)

Eliminate prime notation - use Leibnitz' powerful d

This is not just my personal comment. Prime notation for derivatives is:

${\displaystyle f'=\mathrm {effing\,stupid} \,}$

because:

• its easy to loose this when writing for a dot, splodge, smudge etc. and may lead to errors,
• it is unclear what to differentiate with respect to, context needs to provide explanation instead of stating the variable. With d notation only the variables and their symbols need to be stated - then its absolutely clear what dy, dy/dx is, the infinitesimal amount simply has a d in front, using subscripts just clutters notation, which should be used for index variables or naming variables,
• more intuitive: this is not actually what d is, but is a correct interpretation: d can be considered as a very very very (...) small number tending asymptotically to zero:

${\displaystyle \mathrm {d} x=\lim _{\delta x\rightarrow 0}\delta x\,}$

"Multiplying" by another d gives an even smaller quantity: :${\displaystyle \mathrm {d} (\mathrm {d} x)=\mathrm {dd} x=\mathrm {d} ^{2}x\,}$

"Multiplying" by d any number of times gives an even smaller still quantity: ${\displaystyle \mathrm {d} (\cdots (\mathrm {d} (\mathrm {d} x)))=\mathrm {d\cdots dd} x=\mathrm {d} ^{n}x\,}$

We can also multiply the full dx with itself as many times as needed, raising it to a power:

${\displaystyle (\mathrm {d} x)(\mathrm {d} x)\cdots (\mathrm {d} x)=\mathrm {d} x\mathrm {d} x\cdots \mathrm {d} x=(\mathrm {d} x)^{n}=\mathrm {d} x^{n}\,}$

this explains the notation:

${\displaystyle {\frac {\mathrm {d} ^{n}y}{\mathrm {d} x^{n}}}={\frac {\mathrm {d} }{\mathrm {d} x}}\left(\cdots {\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)\right)={\frac {\mathrm {d} \cdots \mathrm {d} \mathrm {d} y}{\mathrm {d} x\cdots \mathrm {d} x\mathrm {d} x}}={\frac {\mathrm {d} ^{n}y}{\mathrm {d} x^{n}}}}$

• the notion of an infinitesimal quantity is too powerful to ignore. The ratio of two infinitesimal quantities (say) dy/dx is either that or the derivative of y w.r.t. x. The manipulation of multiplying/dividing by infinitesimal quantities can be just that and allow all derivatives to be derived. To derive the ratio of infinitesimal amounts of variables y as a function of x:

${\displaystyle \mathrm {d} x=\lim _{\delta x\to 0}\delta x\,}$

left-hand dy

${\displaystyle \mathrm {d} y=\lim _{\delta y\to 0}\delta y=\lim _{\delta x\to 0^{+}}\,}$

right-hand dy

${\displaystyle \mathrm {d} y=\lim _{\delta y\to 0}\delta y=\lim _{\delta x\to 0^{-}}\,}$

left and right dy must be equal for continuity, the right-hand ratio of two quantities (also a derivative)

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {\lim _{\delta y\to 0^{+}}\delta y}{\lim _{\delta x\to 0}\delta x}}={\frac {\lim _{\delta x\to 0^{+}}}{\lim _{\delta x\to 0}\delta x}}=\lim _{\delta x\to 0^{+}}\left\,}$

left-hand ratio of two quantities (again also a derivative)

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {\lim _{\delta y\to 0^{-}}\delta y}{\lim _{\delta x\to 0}\delta x}}={\frac {\lim _{\delta x\to 0^{-}}}{\lim _{\delta x\to 0}\delta x}}=\lim _{\delta x\to 0^{-}}\left\,}$

again left and right dy/dx must be equal for continuity. Prime notation hides this: the ratio of two infinitesimal quantities is more general than a derivative. dy and dy x can be treated as algebraic quantities. Ex. the length of a curve L (in 2d Cartesian plane) can be written directly as Pythagoras' theorem:

${\displaystyle \mathrm {d} \ell ^{2}=\mathrm {d} x^{2}+\mathrm {d} y^{2}\,}$

${\displaystyle \mathrm {d} \ell ={\sqrt {\mathrm {d} x^{2}+\mathrm {d} y^{2}}}\,}$

${\displaystyle \mathrm {d} \ell ={\sqrt {\mathrm {d} x^{2}+{\frac {\mathrm {d} x^{2}}{\mathrm {d} x^{2}}}\mathrm {d} y^{2}}}\,}$

${\displaystyle \mathrm {d} \ell =\mathrm {d} x{\sqrt {1+{\frac {\mathrm {d} y^{2}}{\mathrm {d} x^{2}}}}}\,}$

${\displaystyle L=\int \mathrm {d} \ell =\int {\sqrt {1+{\frac {\mathrm {d} y^{2}}{\mathrm {d} x^{2}}}}}\mathrm {d} x\,}$

• It makes integration much easier, to see which infinitesimal quantities cancel when integrating a derivative. It makes integration by parts and by substitution trillions of times more obvious and easier.

The differential product rule is clear:

${\displaystyle \mathrm {d} (pq)=p\mathrm {d} q+q\mathrm {d} p\,}$

${\displaystyle {\frac {\mathrm {d} (pq)}{\mathrm {d} t}}={\frac {p\mathrm {d} q+q\mathrm {d} p}{\mathrm {d} t}}={\frac {p\mathrm {d} q}{\mathrm {d} t}}+{\frac {q\mathrm {d} p}{\mathrm {d} t}}\,}$

hence integration by parts is clear:

${\displaystyle \int {\frac {\mathrm {d} (pq)}{\mathrm {d} t}}\mathrm {d} t=\int p\mathrm {d} q+\int q\mathrm {d} p=\int {\frac {p\mathrm {d} q}{\mathrm {d} t}}\mathrm {d} t+\int {\frac {q\mathrm {d} p}{\mathrm {d} t}}\mathrm {d} t\,\!}$

${\displaystyle \int \mathrm {d} (pq)=\int p\mathrm {d} q+q\mathrm {d} p=\int p\mathrm {d} q+\int q\mathrm {d} p\,}$

${\displaystyle pq=\int p\mathrm {d} q+\int q\mathrm {d} p\,}$

whatever p or q are, if they can quickly be tucked under the d, integration by parts is really quick.

Integration by sub is obvious:

${\displaystyle y=y(x),x=x(t)\,}$

${\displaystyle \mathrm {d} x={\frac {\mathrm {d} x}{\mathrm {d} t}}\mathrm {d} t\,}$

${\displaystyle \int y(x)\mathrm {d} x=\int y(x(t)){\frac {\mathrm {d} x(t)}{\mathrm {d} t}}\mathrm {d} t\,}$

Just look at the notation alive in action - it’s on the loose, dynamic, manipulating, resembles the familiar elementary algebra, and implies a direct formalism with linear algebra...

As such I'm going to remove all prime notation in this article and use the almighty tools of Leibnitz: the d, and the ∫. Anyone reading this will think I am being sooo pedantic and conscious about notation. But consciousness of powerful notation is a good thing.

Maschen (talk) 18:51, 21 November 2011 (UTC)

Wow - no one has objected after a week. I'll go for it. Also

• i'm going cut down the monotonous repetetive writing in each subsection (e.x. "for functions f and g", "for functions f and g", "for functions f and g" .....) by collecting base formulae (product/quotient/chain/general power rulse etc) together,
• write the definitions of the function in the standard formalism (e.x.):

${\displaystyle f:\mathbf {R} \rightarrow \mathbf {R} \,\!}$

to clarify what the function must be for the derivative to exist. E.x. cannot have a function f(x) which vanishes at x₀ (i.e. f(x₀)=0 ) and then calculate the diff of 1/f(x₀), so the mapping would be:

${\displaystyle f:\mathbf {R} -\{x_{0}\}\rightarrow \mathbf {R} \,\!}$

for for composition f(g(x)):

${\displaystyle f\circ g:\mathbf {R} \rightarrow \mathbf {R} \,\!}$

(I h8 this circle notation - it also looks stupid.....).

I suppose someone will still merrily delete my edits after, even with this much justification for an effort to clarify, not complicate. Please at least see where i'm coming from before doing so.--Maschen (talk) 17:42, 1 December 2011 (UTC)

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