Question Number 157784 by Dhetal last updated on 27/Oct/21 $${from}\:{the}\:{partical}\:{differential}\:{equation}\:{by}\:{eliminating}\:{constants}\:{indicated}\:{in}\:{brackets}\:{from}\:{the}\:{following}\:{equation}\:{z}=\left({x}+{a}\right)\:\left({y}+{b}\right);\left({a},{b}\right) \\ $$ Commented by Tawa11 last updated on 27/Oct/21 $$\mathrm{Great}\:\mathrm{sir} \\ $$ Commented by Ali_Adily…
Question Number 91911 by jagoll last updated on 03/May/20 $$\mathrm{y}''''+\mathrm{2y}''+\mathrm{y}=\mathrm{sin}\:\mathrm{x}\: \\ $$ Answered by niroj last updated on 03/May/20 $$\:\:\mathrm{y}^{''''} +\mathrm{2y}^{''} +\mathrm{y}=\:\mathrm{sin}\:\mathrm{x} \\ $$$$\:\:\:\frac{\mathrm{d}^{\mathrm{4}} \mathrm{y}}{\mathrm{dx}^{\mathrm{4}}…
Question Number 26368 by d.monhbayr@gmail.com last updated on 24/Dec/17 $${y}={a}^{\mathrm{arc}{tg}\sqrt{{x}}} \\ $$$${derivative}\:? \\ $$ Commented by kaivan.ahmadi last updated on 24/Dec/17 $$\mathrm{log}_{\mathrm{a}} \mathrm{y}=\mathrm{arctg}\sqrt{\mathrm{x}}\Rightarrow\frac{\mathrm{y}'}{\mathrm{y}}\mathrm{lna}=\frac{\left(\sqrt{\mathrm{x}}\right)^{'} }{\mathrm{1}+\left(\sqrt{\mathrm{x}}\right)^{\mathrm{2}} }=…
Question Number 26365 by d.monhbayr@gmail.com last updated on 24/Dec/17 $${y}=\mathrm{log}_{{a}} \left({x}^{\mathrm{2}} −\mathrm{16}\right) \\ $$ Commented by mrW1 last updated on 24/Dec/17 $${what}\:{is}\:{your}\:{question}? \\ $$ Terms…
Question Number 26364 by d.monhbayr@gmail.com last updated on 24/Dec/17 $${y}={x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${min}=?\:\:\:{max}=? \\ $$$${help}\:{pls} \\ $$ Commented by mrW1 last updated on 24/Dec/17 $${there}\:{is}\:{no}\:{max},\:{since}\:{y}\rightarrow\infty\:{when}\:{x}\rightarrow\pm\infty.…
Question Number 26363 by d.monhbayr@gmail.com last updated on 24/Dec/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+{x}\mathrm{sin}\:{x}}−\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}{{tg}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$ Commented by kaivan.ahmadi last updated on 24/Dec/17 $$=\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−\mathrm{2x}^{\mathrm{2}} }}{\frac{\mathrm{x}^{\mathrm{2}}…
Question Number 91848 by jagoll last updated on 03/May/20 $${y}^{''} −\mathrm{4}{y}^{'} +\mathrm{4}{y}=\left({x}+\mathrm{1}\right){e}^{\mathrm{2}{x}} \\ $$$$ \\ $$ Commented by john santu last updated on 03/May/20 $${complementary}\:{solution}…
Question Number 26299 by d.monhbayr@gmail.com last updated on 23/Dec/17 $${y}=\frac{\mathrm{1}+{e}^{{x}} }{\mathrm{1}−{e}^{{x}} } \\ $$$${y}'=? \\ $$ Commented by abdo imad last updated on 24/Dec/17 $$\frac{{dy}}{{dx}}=\frac{{d}\left(\mathrm{1}+{e}^{{x}}…
Question Number 26300 by d.monhbayr@gmail.com last updated on 23/Dec/17 $${y}={a}^{\mathrm{arc}{tg}\sqrt{{x}}} \\ $$$${y}'=? \\ $$ Commented by abdo imad last updated on 24/Dec/17 $${we}\:{have}\:\:{y}=\:{e}^{{arctan}\left(\sqrt{\left.{x}\right)}{lna}\:\right.} \:\:\:{with}\:{condition}\:{a}>\mathrm{0} \\…
Question Number 26297 by d.monhbayr@gmail.com last updated on 23/Dec/17 $${y}={x}−\frac{\mathrm{2}}{{x}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} } \\ $$$${y}'=? \\ $$ Answered by Joel578 last updated on 24/Dec/17 $$\:{y}\:=\:{x}\:−\:\mathrm{2}{x}^{−\mathrm{4}} \:−\:\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\mathrm{3}}…