Question Number 103870 by Ar Brandon last updated on 17/Jul/20 $$\mathcal{D}\acute {\mathrm{e}montrer}\:\mathrm{que}\:\mathrm{la}\:\mathrm{fonction}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \centerdot\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:\mathrm{admet}\:\mathrm{un}\:\mathrm{DL}\:\mathrm{d}'\mathrm{ordre}\:\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 103828 by ~blr237~ last updated on 17/Jul/20 $$\:\:\:\:\:{min}\left\{\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} −{px}−{q}\right)^{\mathrm{2}} {dx}\:,\:\:\left({p},{q}\right)\in\mathbb{R}^{\mathrm{2}} \:\right\} \\ $$ Answered by bobhans last updated on 17/Jul/20 $$\left({x}^{\mathrm{3}}…
Question Number 103820 by ~blr237~ last updated on 17/Jul/20 $$\:\:\:\int_{\mathbb{R}} ^{} \:\frac{{e}^{−\mathrm{2}{i}\pi{ax}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=\:\pi{e}^{−\mathrm{2}\pi{a}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\pi{a}\right)\:\:\:\:\:\:\:\:\:{a}>\mathrm{0} \\ $$ Answered by mathmax by abdo last updated…
Question Number 38232 by rahul 19 last updated on 23/Jun/18 $$\mathrm{Differentiate}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}}{{x}}\right)\:\: \\ $$$${without}\:{using}\:{any}\:{trigonometric}\: \\ $$$${substitution}\:! \\ $$ Commented by math khazana…
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}}…