Question Number 167336 by mnjuly1970 last updated on 13/Mar/22 $$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{0}<\:{x}<\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{f}\left({x}\right)=\:\sqrt[{\mathrm{20}}]{{sin}\left({x}\right)\:\:\:}\:+\:\sqrt[{\mathrm{20}}]{{cos}\left({x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:{R}_{\:{f}} \:=?\:\:\left({Range}\:\right) \\ $$$$ \\ $$ Answered by MJS_new last…
Question Number 167321 by infinityaction last updated on 13/Mar/22 Commented by mr W last updated on 13/Mar/22 $${what}'{s}\:{the}\:{question}? \\ $$ Commented by infinityaction last updated…
Question Number 167301 by mnjuly1970 last updated on 12/Mar/22 Answered by amin96 last updated on 12/Mar/22 $$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}−\boldsymbol{\mathrm{cos}}\left(\frac{\pi}{\mathrm{4}}−\boldsymbol{\mathrm{x}}\right)\right)}\boldsymbol{\mathrm{dx}}\:\:\:\:\frac{\pi}{\mathrm{4}}−\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{t}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{t}}\left[\frac{\pi}{\mathrm{8}};\:\frac{\pi}{\mathrm{4}}\right]\:\:\:\:\boldsymbol{{I}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\frac{\pi}{\mathrm{8}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\boldsymbol{\mathrm{dt}}}{\left(\mathrm{1}−\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{t}}\right)\right)}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\frac{\pi}{\mathrm{8}}}…
Question Number 167267 by mnjuly1970 last updated on 11/Mar/22 Answered by mindispower last updated on 15/Mar/22 $${IBP}\Rightarrow\Omega=\left[{ln}\left(\mathrm{1}+{x}\right){Li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right){ln}\left({x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 36178 by prof Abdo imad last updated on 30/May/18 $${let}\:{f}\left({x},{y}\right)={ln}\left(\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }\right)\: \\ $$$${calculate}\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }\left({x},{y}\right)+\frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} } \\ $$ Commented by abdo…
Question Number 36177 by prof Abdo imad last updated on 30/May/18 $${let}\:{f}\left({x}\right)=\:{arctan}\left(\frac{{x}}{{y}}\right) \\ $$$${calculate}\:\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }\left({x},{y}\right)\:,\:\frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }\left({x},{y}\right),\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}\left({x},{y}\right) \\ $$$$\frac{\partial^{\mathrm{2}} {f}}{\partial{y}\partial{x}}\left({x},{y}\right) \\ $$ Commented…
Question Number 36174 by prof Abdo imad last updated on 30/May/18 $${find}\:{lim}_{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} \:\:\:\frac{\mathrm{1}−{cos}\left(\sqrt{{xy}}\right)}{{y}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 36175 by prof Abdo imad last updated on 30/May/18 $${let}\:{g}\left({x},{y}\right)\:=\:\frac{\mathrm{1}+{x}+{y}}{{x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} } \\ $$$${is}\:{g}\:{have}\:{a}\:{limit}\:{at}\:\left(\mathrm{0},\mathrm{0}\right)? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 36176 by prof Abdo imad last updated on 30/May/18 $${let}\:{f}\left({x},{y}\right)\:=\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){sin}\left\{\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}\right\}\:{if}\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right) \\ $$$${and}\:{f}\left(\mathrm{0},\mathrm{0}\right)=\mathrm{0} \\ $$$${prove}\:{that}\:{f}\:{is}\:{differenciable}\:{at}\:{all}\:{point}\:{of}\:{R}^{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\frac{\partial{f}}{\partial{x}}\:{and}\:\frac{\partial{f}}{\partial{y}}\:{are}\:{not}\:{differdnciable} \\ $$$${at}\:\left(\mathrm{0},\mathrm{0}\right) \\…
Question Number 36173 by prof Abdo imad last updated on 30/May/18 $${calculate}\:\frac{\partial{f}}{\partial{x}}\:{and}\:\frac{\partial{f}}{\partial{y}}\:{in}\:{this}\:{cases} \\ $$$$\left.\mathrm{1}\right)\:{f}\left({x},{y}\right)=\:{e}^{−{x}} \:{sin}\left(\mathrm{2}{y}\:+\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right){f}\left({x},{y}\right)\:=\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){e}^{−{xy}} \\ $$$$\left.\mathrm{3}\right){f}\left({x},{y}\right)\:=\:\frac{{x}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} } \\ $$…