# let-S-i-a-n-x-i-prove-or-disprove-that-d-dx-i-a-n-x-i-i-a-n-d-dx-x-i-i-a-n-x-i-

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