Question Number 164366 by Zaynal last updated on 16/Jan/22 $$\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\:\left(\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \right) \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{\mathrm{A}}\right\} \\ $$ Commented by cortano1 last updated on 16/Jan/22 $$\:{y}\:=\:{e}^{\mathrm{tan}\:{x}} \\ $$$$\:\mathrm{ln}\:{y}\:=\:\mathrm{tan}\:{x}…
Question Number 164339 by mnjuly1970 last updated on 16/Jan/22 $$ \\ $$$$\:\:\:\:{In}\:\:{A}\overset{\Delta} {{B}C}\:\:\::\:\:\:{cos}^{\:\mathrm{2}} \left({A}\:\right)+\:{cos}^{\:\mathrm{2}} \left({B}\:\right)+\:{cos}^{\:\mathrm{2}} \left(\:{C}\:\right)=\mathrm{1}\:\:. \\ $$$$\:\:\:\:\:\:\:\:{Prove}\:{that}\:\:{A}\overset{\Delta} {{B}C}\:\:\:{is}\:\:\:{right}\:{angled}. \\ $$$$\:\:\:\:\:\:−−−−−−−− \\ $$$$\:\:\:\:\:\: \\ $$…
Question Number 33208 by abdo imad last updated on 12/Apr/18 $${solve}\:{the}\:{d}.{e}.\:{x}^{''} \left({t}\right)\:+\mathrm{3}{x}^{'} \left({t}\right)\:+\mathrm{2}\:{x}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{e}^{{t}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33209 by abdo imad last updated on 12/Apr/18 $${solve}\:{the}\:{system}\:\:{x}^{'} \:={ay}\:{and}\:{y}^{'} \:=−{ax}\:.\:{afrom}\:{R}\:\:,{a}\neq\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33122 by abdo imad last updated on 10/Apr/18 $${let}\:{considere}\:{f}\:{and}\:{u}\:{differenciable}\:{function}\:{prove} \\ $$$${that}\:\frac{{d}}{{dt}}\left(\:\int_{{a}} ^{{u}\left({t}\right)} {f}\left({t},{x}\right){dx}\right)=\int_{{a}} ^{{u}\left({t}\right)} \:\frac{\partial{f}}{\partial{t}}\left({t},{x}\right){dx}\:+{f}\left({t},{u}\left({t}\right)\right){u}^{'} \left({t}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 164103 by mnjuly1970 last updated on 14/Jan/22 $$ \\ $$$$\:\:\:{prove}\:{that} \\ $$$$\: \\ $$$$\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\:\right).\frac{{dx}}{{x}\:\sqrt{\:\mathrm{1}−{x}^{\:\mathrm{2}} }}\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:−−\:{m}.{n}−− \\ $$$$ \\…
Question Number 164059 by bobhans last updated on 13/Jan/22 $$\:\:\mathrm{Air}\:\mathrm{leaks}\:\mathrm{from}\:\mathrm{a}\:\mathrm{spherical}\:\mathrm{ballon}\:\mathrm{so}\:\mathrm{that}\: \\ $$$$\:\mathrm{it}\:\mathrm{maintains}\:\mathrm{its}\:\mathrm{shape}\:\mathrm{at}\:\mathrm{a}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{25}\:\mathrm{cc}/\mathrm{m} \\ $$$$\:.\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{change}\:\mathrm{in}\:\mathrm{the}\:\mathrm{length} \\ $$$$\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{balloon}\:\mathrm{when}\:\mathrm{the}\:\mathrm{radius} \\ $$$$\:\:\mathrm{is}\:\mathrm{5}\:\mathrm{cm} \\ $$ Terms of Service Privacy Policy…
Question Number 32980 by Nayon.Sm last updated on 08/Apr/18 $$ \\ $$$$ \\ $$$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\begin{vmatrix}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{x}^{\mathrm{3}} +\mathrm{3}} }&{\mathrm{2}^{\mathrm{x}} }&{\mathrm{cosx}^{\mathrm{x}} }\\{\mathrm{log}_{\mathrm{2}^{\mathrm{x}+\mathrm{1}} } \left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2x}} +\overset{} {\mathrm{3}}\right)}&{\mathrm{xlnx}}&{\mathrm{sin}^{−\mathrm{1}} \mathrm{tanx}}\\{\mathrm{3}}&{\pi^{\mathrm{sinhx}}…
Question Number 32971 by math1967 last updated on 08/Apr/18 $${If}\:{siny}={xsin}\left({a}+{y}\right)\:{show}\:{that} \\ $$$$\frac{{dy}}{{dx}}=\frac{{sina}}{\mathrm{1}−\mathrm{2}{xcosa}+{x}^{\mathrm{2}} } \\ $$$$ \\ $$ Answered by math1967 last updated on 06/May/18 $${siny}={xsinacosy}\:+{xcosasiny}…
Question Number 98450 by bemath last updated on 14/Jun/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{supremum}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{infimum}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}\:,\mathrm{x}\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\:\right] \\ $$ Commented by bobhans last updated on 14/Jun/20 $$\mathrm{f}\:'\left(\mathrm{x}\right)=\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{xcos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:\:\_\_\_\left(\mathrm{1}\right) \\ $$$$\mathrm{take}\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\mathrm{sin}\:\mathrm{x}−\mathrm{xcos}\:\mathrm{x}\:;\:\mathrm{x}\in\:\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\:\right]\:…