Question Number 116319 by bemath last updated on 03/Oct/20 $$\mathrm{If}\:\mathrm{z}\:=\:\mathrm{x}^{\mathrm{2}} \:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{y}}{\mathrm{x}}\right),\:\mathrm{find}\:\frac{\partial^{\mathrm{2}} \mathrm{z}}{\partial\mathrm{x}\partial\mathrm{y}}\: \\ $$$$\mathrm{at}\:\left(\mathrm{1},\mathrm{1}\right) \\ $$ Answered by john santu last updated on 03/Oct/20…
Question Number 50752 by Tinkutara last updated on 19/Dec/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 116284 by mnjuly1970 last updated on 02/Oct/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…\:\:{calculus}\:{I}\:… \\ $$$$\:\:\:\:\:{evaluate}\::\: \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{I}\::=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{3}}} {log}\left(\mathrm{1}+\sqrt{\mathrm{3}}\:{tan}\left({x}\right)\right){dx}=??? \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by Dwaipayan…
Question Number 50430 by Abdo msup. last updated on 16/Dec/18 $${solve}\:\left({x}^{\mathrm{2}} \:+\mathrm{3}\right){y}^{'} \:+\left({x}^{\mathrm{3}} −\mathrm{1}\right){y}\:={x}^{\mathrm{2}} \\ $$ Commented by Abdo msup. last updated on 17/Dec/18 $$\left({he}\right)\:\:\left({x}^{\mathrm{2}}…
Question Number 50431 by Abdo msup. last updated on 16/Dec/18 $${solve}\:{xy}^{'} \:+{y}\:=\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }} \\ $$ Commented by Abdo msup. last updated on 16/Dec/18 $$\left({he}\right)\:\:\:\:\:{xy}^{'} \:+{y}\:=\mathrm{0}\:\Rightarrow{xy}^{'}…
Question Number 50432 by Abdo msup. last updated on 16/Dec/18 $${solve}\:{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}^{'} \:+\mathrm{2}{y}\:={x}^{\mathrm{2}} \\ $$ Commented by Abdo msup. last updated on 16/Dec/18 $$\left({he}\right)\:\Rightarrow{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}^{'}…
Question Number 50428 by Abdo msup. last updated on 16/Dec/18 $${solve}\:{y}^{'} \:+\frac{\mathrm{2}{x}+\mathrm{1}}{{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:{y}\:=\:\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 50429 by Abdo msup. last updated on 16/Dec/18 $${solve}\:{y}^{''} \:+{e}^{{x}^{\mathrm{2}} } {y}\:=\mathrm{0} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 19/Dec/18 $${prof}\:{Abdo}\:{it}\:{is}\:{really}\:{a}\:{excellent}\:{problem}…{still} \\…
Question Number 115750 by shahria14 last updated on 28/Sep/20 Answered by Dwaipayan Shikari last updated on 28/Sep/20 $$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{log}\left(\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \right)}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$$\mathrm{log}\left(\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \right)=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{approx}\right)…
Question Number 115733 by bemath last updated on 28/Sep/20 $${Given}\:{xy}\:=\frac{\mathrm{1}}{\mathrm{3}}\:{for}\:{x},{y}\in\mathbb{R}\: \\ $$$${find}\:{min}\:{value}\:{of}\:\frac{\mathrm{4}}{{x}^{\mathrm{6}} }+\frac{\mathrm{9}}{{y}^{\mathrm{6}} } \\ $$ Answered by MJS_new last updated on 28/Sep/20 $${y}=\frac{\mathrm{1}}{\mathrm{3}{x}}\:\Rightarrow\:\frac{\mathrm{4}}{{x}^{\mathrm{6}} }+\frac{\mathrm{9}}{{y}^{\mathrm{6}}…