Question Number 73566 by Rio Michael last updated on 13/Nov/19
$${show}\:{that}\:{f}\left({x}\right)\:=\:\mid{x}\mid\:{is}\:{not}\:{differentiable}\:{at}\:{x}=\mathrm{0},\:{where}\:\mid{x}\mid \\ $$$${denotes}\:{he}\:{absolute}\:{value}\:{function} \\ $$
Commented by Rio Michael last updated on 13/Nov/19
$${thanks}\:{sir}, \\ $$$${but}\:{when}\:{looking}\:{at}\:{differentiability} \\ $$$${should}\:{we}\:{take}\: \\ $$$${derivatives}\:{at}\:{x}\:=\:\mathrm{0}\:{from}\:{both}\:{direction}? \\ $$
Commented by kaivan.ahmadi last updated on 13/Nov/19
$${f}'_{−} \left(\mathrm{0}\right)=−\mathrm{1} \\ $$$${f}'_{+} \left(\mathrm{0}\right)=\mathrm{1} \\ $$$$ \\ $$
Commented by malwaan last updated on 14/Nov/19
$${yes} \\ $$