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Question Number 3385 by Filup last updated on 12/Dec/15
let: S=Σ_(i=a) ^n x_i prove or disprove that: (d/dx)(Σ_(i=a) ^n x_i )=Σ_(i=a) ^n ((d/dx){x_i }) =Σ_(i=a) ^n x_i ^′
$$\mathrm{let}: \\ $$$$\:\:\:\:\:\:\:\:{S}=\underset{{i}={a}} {\overset{{n}} {\sum}}{x}_{{i}} \\ $$$$\mathrm{prove}\:\mathrm{or}\:\mathrm{disprove}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\:\frac{{d}}{{dx}}\left(\underset{{i}={a}} {\overset{{n}} {\sum}}{x}_{{i}} \right)=\underset{{i}={a}} {\overset{{n}} {\sum}}\left(\frac{{d}}{{dx}}\left\{{x}_{{i}} \right\}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{i}={a}} {\overset{{n}} {\sum}}{x}_{{i}} ^{'} \\ $$
Commented by prakash jain last updated on 12/Dec/15
Some of two function which are not differentiable at x=x_0 may be differentiable at x=x_0 .
$$\mathrm{Some}\:\mathrm{of}\:\mathrm{two}\:\mathrm{function}\:\mathrm{which}\:\mathrm{are}\:\mathrm{not}\:\mathrm{differentiable} \\ $$$$\mathrm{at}\:{x}={x}_{\mathrm{0}} \:\mathrm{may}\:\mathrm{be}\:\mathrm{differentiable}\:\mathrm{at}\:{x}={x}_{\mathrm{0}} . \\ $$
Commented by Filup last updated on 12/Dec/15
i was meant to write that x_i is the i^(th) term in a series
$${i}\:{was}\:{meant}\:{to}\:{write}\:{that} \\ $$$${x}_{{i}} \:{is}\:{the}\:{i}^{\mathrm{th}} \:{term}\:{in}\:{a}\:{series} \\ $$
Commented by prakash jain last updated on 12/Dec/15
It does make a difference. If x_i i^(th) term in series. It is a still function of x.
$$\mathrm{It}\:\mathrm{does}\:\mathrm{make}\:\mathrm{a}\:\mathrm{difference}.\:\mathrm{If}\:{x}_{{i}} \:{i}^{{th}} \:\mathrm{term}\:\mathrm{in} \\ $$$$\mathrm{series}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{still}\:\mathrm{function}\:\mathrm{of}\:{x}. \\ $$
Commented by Filup last updated on 12/Dec/15
i see. hmm... so is there ever a time such that the above is true? other than x_i =constant
$$\mathrm{i}\:\mathrm{see}.\:\mathrm{hmm}…\:\mathrm{so}\:\mathrm{is}\:\mathrm{there}\:\mathrm{ever}\:\mathrm{a}\:\mathrm{time} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{above}\:\mathrm{is}\:\mathrm{true}? \\ $$$$\mathrm{other}\:\mathrm{than}\:{x}_{{i}} =\mathrm{constant} \\ $$
Commented by prakash jain last updated on 12/Dec/15
It is generally true if each of x_i is differentiable. I gave an exception rather than rule.
$$\mathrm{It}\:\mathrm{is}\:\mathrm{generally}\:\mathrm{true}\:\mathrm{if}\:\mathrm{each}\:\mathrm{of}\:{x}_{{i}} \:\mathrm{is}\:\mathrm{differentiable}. \\ $$$$\mathrm{I}\:\mathrm{gave}\:\mathrm{an}\:\mathrm{exception}\:\mathrm{rather}\:\mathrm{than}\:\mathrm{rule}. \\ $$
Commented by Filup last updated on 12/Dec/15
Ah I understand now, thanks!
$$\mathrm{Ah}\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now},\:\mathrm{thanks}! \\ $$
Commented by prakash jain last updated on 12/Dec/15
(d/dx)(f(x)+g(x))=(d/dx)f(x)+(d/dx)g(x) what you have above is a essentially same formula.
$$\frac{{d}}{{dx}}\left({f}\left({x}\right)+{g}\left({x}\right)\right)=\frac{{d}}{{dx}}{f}\left({x}\right)+\frac{{d}}{{dx}}{g}\left({x}\right) \\ $$$${what}\:{you}\:{have}\:{above}\:{is}\:{a}\:{essentially}\:{same} \\ $$$${formula}. \\ $$
Answered by prakash jain last updated on 12/Dec/15
x_1 = { (2,(x≥1)),(3,(x<1)) :} x_2 = { ((−2),(x≥1)),((−3),(x<1)) :} (d/dx)(x_1 +x_2 ) at x=1 is not equal to (d/dx)x_1 +(d/dx)x_2 at x=1.
$${x}_{\mathrm{1}} =\begin{cases}{\mathrm{2}}&{{x}\geqslant\mathrm{1}}\\{\mathrm{3}}&{{x}<\mathrm{1}}\end{cases} \\ $$$${x}_{\mathrm{2}} =\begin{cases}{−\mathrm{2}}&{{x}\geqslant\mathrm{1}}\\{−\mathrm{3}}&{{x}<\mathrm{1}}\end{cases} \\ $$$$\frac{{d}}{{dx}}\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)\:\mathrm{at}\:{x}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\frac{{d}}{{dx}}{x}_{\mathrm{1}} +\frac{{d}}{{dx}}{x}_{\mathrm{2}} \\ $$$$\mathrm{at}\:{x}=\mathrm{1}. \\ $$

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