Question Number 171267 by pete last updated on 11/Jun/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{y}=\frac{{x}}{\mathrm{1}+{x}}\:\mathrm{at}\:\mathrm{which}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{are}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:{x}−{y}+\mathrm{8}=\mathrm{0}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at} \\ $$$$\mathrm{these}\:\mathrm{points}. \\ $$ Commented by greougoury555 last updated…
Question Number 171265 by ali009 last updated on 11/Jun/22 $${find}\:{the}\:{drivative}\:{of}\: \\ $$$${f}\left({x},{y},{z}\right)={cos}\left({xy}\right)+{e}^{{zy}} +{ln}\left({zy}\right) \\ $$$${at}\:{point}\:\left(\mathrm{1},\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right)\:{in}\:{the}\:{direction} \\ $$$${v}={i}+\mathrm{2}{j}+\mathrm{2}{k} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 105706 by bramlex last updated on 31/Jul/20 $$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{dx}}{\left(\sqrt{\mathrm{5}}−\mathrm{cos}\:{x}\right)^{\mathrm{3}} }\:? \\ $$ Answered by john santu last updated on 31/Jul/20 $${F}\left({a}\right)\:=\:\underset{\mathrm{0}} {\overset{\pi}…
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Question Number 40103 by maxmathsup by imad last updated on 15/Jul/18 $${let}\:{g}\left({x}\right)=\sqrt{−{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{g}\:{is}\:{solution}\:{for}\:{the}\:{differencial}\:{equation} \\ $$$$\mathrm{4}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} \:+\mathrm{4}{xy}^{'} \:−{y}\:=\mathrm{0}\:\:\:.{prove}\:{that}\:{g}\:{is}\:{C}^{\infty} {on}\:{R} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{a}\:{relation}\:{between}\:{g}^{\left({n}\right)} \left(\mathrm{0}\right)\:{and}\:{g}^{\left({n}+\mathrm{2}\right)} \left(\mathrm{0}\right)…
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Question Number 40052 by LXZ last updated on 15/Jul/18 $${f}\:'\left({x}\right)={g}\left({x}\right)\:{and}\:\:{g}\:'\left({x}\right)=−{f}\left({x}\right)\:{for} \\ $$$${all}\:{real}\:\:{x}\:\:{andf}\left(\mathrm{5}\right)=\mathrm{2}={f}\:'\left(\mathrm{5}\right)\:{then} \\ $$$${f}^{\mathrm{2}} \left(\mathrm{10}\right)+{g}^{\mathrm{2}} \left(\mathrm{10}\right)\:{is} \\ $$$$\left({a}\right)\:\:\:\mathrm{2}\:\:\:\:\:\left({b}\right)\:\:\:\mathrm{4}\:\:\:\:\:\left({c}\right)\:\:\:\:\mathrm{8}\:\:\:\:\:\left({d}\right)\:{none} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…
Question Number 39930 by Raj Singh last updated on 13/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18 $${one}\:{is}\:{x}\:{and}\:{another}\:{is}\:\mathrm{6}−{x} \\ $$$${y}={x}^{\mathrm{3}} +\left(\mathrm{6}−{x}\right)^{\mathrm{3}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{6}−{x}\right)^{\mathrm{2}} ×−\mathrm{1}…
Question Number 39921 by Raj Singh last updated on 13/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18 $$\left.{iii}\right){max}\:{value}\:{of}\:{sin}\mathrm{2}{x}=\mathrm{1} \\ $$$${min}\:{value}\:{of}\:{sin}\mathrm{2}{x}=−\mathrm{1} \\ $$$${s}\mathrm{0}\:{max}\:{value}\:{ofh}\left({x}\right)=\mathrm{1}+\mathrm{5}=\mathrm{6} \\ $$$${min}\:{value}\:−\mathrm{1}+\mathrm{5}=\mathrm{4} \\…
Question Number 170904 by infinityaction last updated on 03/Jun/22 $$\:\:\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{padel}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{{the}} \\ $$$$\:\:\:\boldsymbol{\mathrm{curve}}\:\:\:\boldsymbol{\mathrm{x}}\:=\: \boldsymbol{{a}\mathrm{cos}\theta}\:−\boldsymbol{\mathrm{acos}}\:\mathrm{2}\boldsymbol{\theta}\:\:, \\ $$$$\:\boldsymbol{\mathrm{y}}\:=\:\mathrm{2}\boldsymbol{\mathrm{a}}\mathrm{sin}\:\boldsymbol{\theta}\:−\boldsymbol{\mathrm{a}}\mathrm{sin}\:\mathrm{2}\boldsymbol{\theta}\:\:\mathrm{is}\:\mathrm{9}\left(\mathrm{r}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)\:=\:\mathrm{8}{p}^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact:…