Question Number 45776 by Sanjarbek last updated on 16/Oct/18
$$\boldsymbol{{y}}=\mid\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\mid \\ $$$$\boldsymbol{{y}}'−? \\ $$
Commented by maxmathsup by imad last updated on 16/Oct/18
$${for}\:{f}\left({x}\right)\neq\mathrm{0}\:{we}\:{have}\:\mid{f}\left({x}\right)\mid=\sqrt{{f}^{\mathrm{2}} \left({x}\right)}\:\Rightarrow\frac{{d}}{{dx}}\left(\mid{f}\left({x}\right)\mid\right)=\frac{\mathrm{2}{f}\left({x}\right){f}^{'} \left({x}\right)}{\mathrm{2}\mid{f}\left({x}\right)\mid}\:=\frac{{f}\left({x}\right){f}^{'} \left({x}\right)}{\mid{f}\left({x}\right)\mid} \\ $$$$=\xi\left({x}\right)\:.{f}^{'} \left({x}\right)\:\:{with}\:\xi\left({x}\right)=\mathrm{1}\:{if}\:{f}\left({x}\right)>\mathrm{0}\:{and}\:\xi\left({x}\right)=−\mathrm{1}\:{if}\:{f}\left({x}\right)<\mathrm{0}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18
$${when}\:{f}\left({x}\right)>\mathrm{0}\:\:\:\:\:\:\mid{f}\left({x}\right)\mid=+{f}\left({x}\right) \\ $$$${when}\:\:{f}\left({x}\right)<\mathrm{0}\:\:\:\:\:\:\mid{f}\left({x}\right)\mid=−{f}\left({x}\right) \\ $$$${so}\:\frac{{dy}}{{dx}}={f}'\left({x}\right)\:\:{f}\left({x}\right)>\mathrm{0} \\ $$$$\left.\frac{{dy}}{{dx}}=−{f}'{x}\right)\:\:{f}\left({x}\right)<\mathrm{0} \\ $$$${so}\frac{{d}\mid{f}\left({x}\right)}{{dx}}\:{doez}\:{notexist}\:\:\:\:\:{pls}\:{check}\:{the}\:{answer} \\ $$