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:…
Question Number 170888 by greougoury555 last updated on 02/Jun/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 105300 by bobhans last updated on 27/Jul/20 $${what}\:{is}\:{the}\:{absolute}\:{speed}\:{of}\:{A}\:{that} \\ $$$${moves}\:{via}\:{y}\:=\:\frac{\mathrm{1}}{\mathrm{1}+{x}}\:\left({upper}\:{part}\right)\:{when} \\ $$$${it}\:{crosses}\:{y}\:−{axis}\:{meanwhile}\:{B} \\ $$$${moves}\:{at}\:{constant}\:{speed}\:{of}\:\mathrm{1}\:{the} \\ $$$${x}−{axis}\:? \\ $$ Terms of Service Privacy Policy…
Question Number 39747 by Raj Singh last updated on 10/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18 $${y}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}{a}−{x}}\:\:\:\frac{{dy}}{{dx}}=\frac{\left(\mathrm{2}{a}−{x}\right)\mathrm{2}{x}−{x}^{\mathrm{2}} \left(\mathrm{0}−\mathrm{1}\right)}{\left(\mathrm{2}{a}−{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{4}{ax}−\mathrm{2}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} }{\left(\mathrm{2}{a}−{x}\right)^{\mathrm{2}}…
Question Number 39709 by Raj Singh last updated on 10/Jul/18 Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18 $${what}\:{is}\:{the}\:{meaning}\:{of}\:{T}^{\mathrm{3}} {T}'\:\:\: \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
Question Number 39582 by Raj Singh last updated on 08/Jul/18 Answered by MrW3 last updated on 08/Jul/18 $${I}\:{think}\:{you}\:{mean}\:{tangents}\:{of}\:{the}\:{curve} \\ $$$${at}\:{positions}\:{where}\:{the}\:{curve}\:{cuts}\:{the} \\ $$$${x}−{axis}. \\ $$$${yx}^{\mathrm{2}} +{x}^{\mathrm{2}}…
Question Number 105113 by yahyajan last updated on 26/Jul/20 Answered by Dwaipayan Shikari last updated on 26/Jul/20 $$\frac{{d}}{{dx}}\left({x}!\right)={y} \\ $$$${logx}+{log}\left({x}−\mathrm{1}\right)+{log}\left({x}−\mathrm{2}\right)+….={logy} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}}+….=\frac{\mathrm{1}}{{y}}\:\frac{{dy}}{{dx}} \\ $$$${x}!\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}}+….\right)=\frac{{dy}}{{dx}} \\…