Question Number 38181 by rahul 19 last updated on 22/Jun/18 $$\mathrm{If}\:\mathrm{y}=\:\:{x}^{\left({lnx}\right)^{{ln}\left({lnx}\right)} } \:{then}\:\frac{{dy}}{{dx}}\:=\:? \\ $$ Commented by rahul 19 last updated on 22/Jun/18 $$\mathrm{I}\:'\mathrm{ve}\:\mathrm{done}\:\mathrm{by}\:\mathrm{taking}\:\mathrm{log}\:\mathrm{and}\:\mathrm{I}'\mathrm{m}\:\mathrm{getting} \\…
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Question Number 38092 by ajfour last updated on 21/Jun/18 Commented by ajfour last updated on 22/Jun/18 $${The}\:{circle}\:{touches}\:{x}=\mathrm{0}\:,\:{y}=\mathrm{0}\:, \\ $$$${and}\:{the}\:{ellipse}\:\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:,\:{in}\:{the} \\ $$$${manner}\:{shown};\:{find}\:{its}\:{radius}\:{R}…
which-is-the-chain-rule-A-dy-dx-dy-dx-1-B-dy-dx-du-dx-dy-dx-C-dy-dx-dy-du-du-dx-D-dy-dx-dy-du-dy-dx-
Question Number 37861 by Rio Mike last updated on 18/Jun/18 $$\:\:\:{which}\:{is}\:{the}\:{chain}\:{rule}? \\ $$$${A}.\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dx}}\:×\:\mathrm{1} \\ $$$${B}.\:\frac{{dy}}{{dx}}\:=\:\frac{{du}}{{dx}}\:×\:\frac{{dy}}{{dx}} \\ $$$${C}.\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{du}}\:×\:\frac{{du}}{{dx}} \\ $$$${D}.\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{du}}\:×\:\frac{{dy}}{{dx}} \\ $$ Commented by ajfour last…
Question Number 37840 by ajfour last updated on 18/Jun/18 $${f}\left(\theta,\phi\right)=\frac{\mathrm{cos}\:\phi\left[\mathrm{cos}\:\theta\:\mathrm{tan}\:\left(\frac{\theta+\phi}{\mathrm{2}}\right)−\mathrm{sin}\:\theta\right]^{\mathrm{2}} }{\mathrm{cos}\:\phi\mathrm{tan}\:\left(\frac{\theta+\phi}{\mathrm{2}}\right)+\mathrm{sin}\:\phi} \\ $$$$\:\phi\:\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right)\:,\:\theta\:\in\:\left(−\frac{\pi}{\mathrm{2}},\:\frac{\pi}{\mathrm{2}}\right); \\ $$$${find}\:{maximum}\:{f}\left(\theta,\phi\right). \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jun/18 $${f}\left(\theta,\phi\right)=\frac{{N}_{{r}}…
Question Number 168905 by safojontoshtemirov last updated on 21/Apr/22 Answered by mindispower last updated on 22/Apr/22 $$\Leftrightarrow{xy}={z};{z}'+\mathrm{2}\frac{{z}^{\mathrm{2}} }{{x}}{ln}\left({x}\right)=\mathrm{0} \\ $$$$−\frac{{z}'}{{z}^{\mathrm{2}} }=\frac{\mathrm{2}{ln}\left({x}\right)}{{x}}\Rightarrow\frac{\mathrm{1}}{{z}}={ln}^{\mathrm{2}} \left({x}\right)+{c}\Rightarrow{z}=\frac{\mathrm{1}}{{ln}^{\mathrm{2}} \left({x}\right)+\mathrm{c}} \\ $$$$\mathrm{y}=\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{ln}^{\mathrm{2}}…
Question Number 168799 by MikeH last updated on 17/Apr/22 $$\mathrm{If}\:\mathrm{the}\:\mathrm{function}\:{f}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{in} \\ $$$$\left[{a},{b}\right]\: \\ $$$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\:\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\frac{{b}−{a}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left({a}+\frac{{k}\left({b}−{a}\right)}{{n}}\right)=\int_{{a}} ^{{b}} {f}\left({x}\right){dx} \\ $$ Terms of…
Question Number 168755 by MikeH last updated on 17/Apr/22 $$\int\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$ Commented by safojontoshtemirov last updated on 17/Apr/22 $$\int\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}−{x}}}{dx}=\:\:\:\sqrt{\mathrm{1}−{x}}={t}\:\:\:\mathrm{1}−{x}={t}^{\mathrm{2}} \:\:{x}=\mathrm{1}−{t}^{\mathrm{2}} \:{dx}=−\mathrm{2}{tdt} \\ $$$$−\int\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} +{t}}{dt}=\int\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}}…
Question Number 103168 by mohammad17 last updated on 13/Jul/20 $${The}\:{particular}\:{solution}\:{of}\:{differential}\:{equation}\: \\ $$$${of}\:\frac{{dy}}{{dx}}+\frac{{y}}{{x}}={k}\:{is}\:{y}=\frac{\mathrm{1}}{{x}}+\mathrm{2}{x}\:{thus}\:,{whats}\:{the}\:{value}\:{of}\:{k}\:? \\ $$ Answered by bemath last updated on 13/Jul/20 $${IF}\:{u}\left({x}\right)=\:{e}^{\int\:\frac{{dx}}{{x}}} \:=\:{e}^{\mathrm{ln}\left({x}\right)} \:=\:{x} \\…
Question Number 103149 by bemath last updated on 13/Jul/20 $${using}\:{first}\:{principal}\: \\ $$$${y}\:=\:\mathrm{ln}\:\left(\mathrm{sin}\:\sqrt{{x}}\right)\:\rightarrow{y}'\:=\:? \\ $$ Answered by bobhans last updated on 13/Jul/20 $${y}'\:=\:\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{sin}\:\sqrt{\mathrm{x}+\bigtriangleup{x}}\right)−\mathrm{ln}\left(\mathrm{sin}\:\sqrt{\mathrm{x}}\right)}{\bigtriangleup{x}} \\ $$$${y}'=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}}…