Question Number 129039 by oustmuchiya@gmail.com last updated on 12/Jan/21 $${partially}\:{Differentiate}\:{function} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)=\mathrm{2}\boldsymbol{{x}}^{−\mathrm{2}} \boldsymbol{{y}}+\boldsymbol{{xy}}^{\mathrm{3}} +\frac{\mathrm{2}}{\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}} \\ $$ Answered by liberty last updated on 12/Jan/21 $$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{2x}^{−\mathrm{2}}…
Question Number 129034 by oustmuchiya@gmail.com last updated on 12/Jan/21 $${Differentiate}\:\boldsymbol{{siny}}\:−\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{3}} −\boldsymbol{{cosx}}=\mathrm{3}\boldsymbol{{y}} \\ $$ Answered by MJS_new last updated on 12/Jan/21 $$\left(\mathrm{cos}\:{y}\:−\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right){dy}+\left(−\mathrm{2}{xy}^{\mathrm{3}} +\mathrm{sin}\:{x}\right){dx}=\mathrm{3}{dy}…
Question Number 63499 by Rio Michael last updated on 04/Jul/19 $${given}\:{that}\:\:\:{a}\mid{b},\:{show}\:{that}\:−{a}\mid{b}. \\ $$ Answered by MJS last updated on 04/Jul/19 $${a}\mid{b}\:\Rightarrow\:\frac{{b}}{{a}}={c}\:\mathrm{with}\:{c}\in\mathbb{Z} \\ $$$$\Rightarrow\:−\frac{{b}}{{a}}=\left(−\mathrm{1}\right)\frac{{b}}{{a}}=\left(−\mathrm{1}\right){c}=−{c} \\ $$$${c}\in\mathbb{Z}\:\Leftrightarrow\:−{c}\in\mathbb{Z}…
Question Number 129033 by oustmuchiya@gmail.com last updated on 12/Jan/21 $${prove}\:{using}\:{the}\:{first}\:{principle}\:{that} \\ $$$${the}\:{derivative}\:{of}\:\boldsymbol{{sin}}\:\boldsymbol{{x}}\:{is}\:\boldsymbol{{cox}}\:\boldsymbol{{x}}\:{and} \\ $$$${that}\:{the}\:{derivative}\:{of}\:\boldsymbol{{cos}}\:\boldsymbol{{x}}\:{is} \\ $$$$−\boldsymbol{{sinx}} \\ $$ Commented by bramlexs22 last updated on 12/Jan/21…
Question Number 129031 by oustmuchiya@gmail.com last updated on 12/Jan/21 $${Given}\:{that}\:\boldsymbol{{tan}}^{−\mathrm{1}} \boldsymbol{{x}}\:{show}\:{that}\:\: \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} } \\ $$ Answered by MJS_new last updated on 12/Jan/21 $${y}=\mathrm{arctan}\:{x} \\…
Question Number 129028 by oustmuchiya@gmail.com last updated on 12/Jan/21 $${Find}\:{gradient}\:{of}\:{the}\:{curve}\:{y}=\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$ Commented by liberty last updated on 12/Jan/21 $$\mathrm{m}\:=\:−\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Terms of…
Question Number 129026 by oustmuchiya@gmail.com last updated on 12/Jan/21 $${differentiate}\:{y}=\frac{\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} \sqrt{\mathrm{6}{x}+\mathrm{2}}}{{x}^{\mathrm{3}} +\mathrm{1}} \\ $$ Answered by MJS_new last updated on 12/Jan/21 $${y}=\frac{{uv}}{{w}} \\ $$$${y}'=\frac{\left({uv}\right)'{w}−{w}'{uv}}{{w}^{\mathrm{2}}…
Question Number 63485 by ANTARES VY last updated on 04/Jul/19 $$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\mathrm{3}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3} \\ $$$$\boldsymbol{\mathrm{F}}\left(\mathrm{2}\right)=\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{F}}\left(−\mathrm{2}\right)=? \\ $$ Answered by MJS last updated on 04/Jul/19 $${f}\left({x}\right)={ax}+{b}…
Question Number 129018 by BHOOPENDRA last updated on 12/Jan/21 $$\frac{\left(\sqrt{{s}}−\mathrm{1}\right)^{\mathrm{2}} }{{s}^{\mathrm{2}} }\:{find}\:{the}\:{inverse}\:{laplace}\:{transformtion} \\ $$ Answered by Dwaipayan Shikari last updated on 12/Jan/21 $$\mathscr{L}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{s}}−\frac{\mathrm{2}}{\:{s}^{\frac{\mathrm{3}}{\mathrm{2}}} }+\frac{\mathrm{1}}{{s}^{\mathrm{2}}…
Question Number 129019 by bramlexs22 last updated on 12/Jan/21 $$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\mathrm{3}\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{4}} +\mathrm{6}\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} + \\ $$$$\:\mathrm{4}\left(\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\right). \\ $$ Answered by MJS_new last updated…