Question Number 171379 by mathlove last updated on 14/Jun/22
$${f}\left({x}\right)=\begin{cases}{\mathrm{2}{x}+\mathrm{3}\:\:\:;{x}>\mathrm{0}}\\{\mathrm{3}{x}−\mathrm{5}\:\:;{x}\leqslant\mathrm{0}}\end{cases} \\ $$ $$\frac{{df}\left({x}\right)}{{dx}}=? \\ $$
Commented byMJS_new last updated on 14/Jun/22
$${f}\left({x}\right)\:=\begin{cases}{\mathrm{3}{x}−\mathrm{5};\:{x}\leqslant\mathrm{0}}\\{\mathrm{not}\:\mathrm{defined};\:\mathrm{0}<{x}\leqslant\mathrm{2}}\\{\mathrm{2}{x}+\mathrm{3};\:{x}>\mathrm{2}}\end{cases} \\ $$ $${f}\:'\left({x}\right)=\begin{cases}{\mathrm{3};\:{x}\leqslant\mathrm{0}}\\{\mathrm{not}\:\mathrm{defined};\:\mathrm{0}<{x}\leqslant\mathrm{2}}\\{\mathrm{2};\:{x}>\mathrm{2}}\end{cases} \\ $$