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? — 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... — 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.