Question Number 52953 by hassentimol last updated on 15/Jan/19
$$ \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{pls}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{that}\:\mathrm{but}\:\mathrm{only} \\ $$$$\mathrm{by}\:\mathrm{using}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of}\:{f}\left(\mathrm{1}\right)\:\mathrm{and}\:\boldsymbol{\mathrm{not}} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{defined}\:\mathrm{defivative}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{itself}\:? \\ $$$$ \\ $$$$\mathrm{Thanks} \\ $$$$ \\ $$$${f}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:−\:{x} \\ $$$$ \\ $$$${f}\:'\:\left(\mathrm{1}\right)\:=\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19
$${f}^{'} \left(\mathrm{1}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left\{\left(\mathrm{1}+{h}\right)^{\mathrm{2}} −\left(\mathrm{1}+{h}\right)\right\}−\left\{\mathrm{1}^{\mathrm{2}} −\mathrm{1}\right\}}{{h}} \\ $$$$\:\:=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{2}{h}+{h}^{\mathrm{2}} −\mathrm{1}−{h}}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{h}\left(\mathrm{1}+{h}\right)}{{h}} \\ $$$$=\mathrm{1}+\mathrm{0}=\mathrm{1}\:\:\:{ans} \\ $$
Commented by hassentimol last updated on 15/Jan/19
$$\:\mathrm{I}\:\mathrm{am}\:\mathrm{so}\:\mathrm{grateful}\:\mathrm{to}\:\mathrm{you}\:!\:\mathrm{I}\:\mathrm{was}\:\mathrm{asking}\:\mathrm{myself} \\ $$$$\:\mathrm{why}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{was}\:\mathrm{incorrect}\:\mathrm{and}\:\mathrm{now},\: \\ $$$$\:\mathrm{thans}\:\mathrm{to}\:\mathrm{you},\:\mathrm{I}\:\mathrm{am}\:\mathrm{understamding}\:! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19
$${f}^{,} \left({a}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({a}+{h}\right)−{f}\left({a}\right)}{{h}}\:\:{this}\:{is}\:{formula} \\ $$$$…{thank}\:{you}… \\ $$