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The Chain Rule for Multivariable Vector-Valued Functions ....
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[QUOTE="fresh_42, post: 6078389, member: 572553"] Which of the following questions does this answer already? I guess this is the question you answered yourself: ##E(y)## is [B]defined[/B] in (17) such that this follows. The general formula for differentiation in direction ##v## goes: ##f\,'(a) \cdot v = f(a+v)-f(a)+r(v)## with a remainder function ##r()## that has to obey ##\lim_{v \to 0} \dfrac{r(v)}{||v||} = 0##. ##E()## is the remainder we get from our formulas. Now we have to show, that the limit condition is obeyed. No. (17) is what defines ##E(y)##, and it depends on both evaluation points. But since ##b=g(a)##, in the end it depends on ##a## alone. But we do not write ##E_a(y)## because we have to distinguish our new remainder function ##E()## from the earlier one ##E_a(y)##. They are two different functions, see (17). We consider the general formula ##f\,'(a) \cdot v = f(a+v)-f(a)+r(v)## again. Now if ##v=0##, we have ##r(v)=0##. Applied to the differentiations of ##f(x)## and ##g(x)## in (14) and (15), this means ##r(0)=E_a(0)=0## resp. ##r(0)=E_b(0)=0##. So by (17) we have ##E(0)= \text{ sth. I }\cdot 0 + \text{ sth. II }\cdot 0##. Now if ##\text{ sth. II }## is bounded as ##y \to 0##, we have a continuous function ##E()## if we [B]set[/B] ##E(0)=0##. Remember, that we defined ##E()## such that it fits our needs. So if the coefficient in (17) at ##E_b(0)## will remain finite if we set ##y=0##, then ##E(0)=0## is a continuous expansion of ##E()## at ##y=0##. We set it so. ##E(y) := \ldots## [/QUOTE]
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The Chain Rule for Multivariable Vector-Valued Functions ....
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