Talk:Leibniz's rule (derivatives and integrals)

Can someone tell me, please, the source from which this proof has been copied? I have some books of analysis and their proofs of this theorem are not so simple. It would be excellent to be sure that we really have an easier proof. Castilla.

I think it would be necesary to proof that uniform continuity allows you to change the limit from the left side of the integral to the right side. I do not see how uniform continuity allows you to state that.

spelling
shouldn't it read Leibniz' rule? &mdash; MFH: Talk 22:39, 10 May 2005 (UTC)


 * Why? Even if his name was Leibnis, there is one convention that dictates to write "Leibnis's", the other being to write "Leibnis'" -- but his name is not spelt Leibnis, it's spelt Leibniz. Dysprosia 03:06, 11 May 2005 (UTC)

Sorry, but I learned it like this: the Genitive s is omitted after the apostrophe if the word ends in an "-s" sound (the reason for the dropped "s" being of course an issue of pronounciation and not of writing). And IMHO, z and tz etc. end in an "-s" sound. Now, you may be right, but I'm curious about how you pronounce this... &mdash; MFH: Talk 20:11, 11 May 2005 (UTC)


 * The /z/ and /tz/ sounds do not end in the softer /s/ sound, I believe. Dysprosia 21:47, 11 May 2005 (UTC)

See http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign too. These all should be one.

The alternate form isn't one
The given alternate form:


 * $$ {d\over dx}\, \int_{f_1(x)}^{f_2(x)} g(t) \,dt = g(f_2(x)) {f_2'(x)} - g(f_1(x)) {f_1'(x)} $$

is merely the fundamental theorem of calculus (FTC) augmented by the chain rule.

Chose any $$ T $$ between $$ {f_1(x)} $$ and $$ {f_2(x)} $$. Then
 * $$ \int_{f_1(x)}^{f_2(x)} g(t) \,dt = \int_{f_1(x)}^{T} g(t) \,dt + \int_{T}^{f_2(x)} g(t) \,dt $$

FTC gives us
 * $$ {d\over dy}\, \int_{y}^{T} g(t) \,dt = -{g(y)} $$ and $$ {d\over dy}\, \int_{T}^{y} g(t) \,dt = {g(y)} $$

So
 * $$ {d\over dx}\, \int_{f_1(x)}^{f_2(x)} g(t) \,dt = {d\over dx}\, (\int_{f_1(x)}^{T} g(t) \,dt + \int_{T}^{f_2(x)} g(t) \,dt ) $$
 * $$ = {d\over dx}\, {\int_{f_1(x)}^{T} g(t) \,dt} + {d\over dx}\, \int_{T}^{f_2(x)} g(t) \,dt = {- g(f_1(x)) {f_1'(x)} + g(f_2(x)) {f_2'(x)}} $$

by the chain rule.