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Category: Differentiation

the-function-is-considered-f-x-y-e-xy-x-y-sen-2x-3y-pi-Calcule-f-x-f-y-2-f-x-2-2-f-x-y-f-x-0-1-f-y-2-1-f-xx-0-1-f-xy-2-1-

Question Number 60426 by cesar.marval.larez@gmail.com last updated on 20/May/19 $${the}\:{function}\:{is}\:{considered}\: \\ $$$${f}\left({x},{y}\right)={e}^{{xy}} +\frac{{x}}{{y}}+{sen}\left(\left(\mathrm{2}{x}+\mathrm{3}{y}\right)\pi\right)\:{Calcule}: \\ $$$$\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}},\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} },\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}.\:\:\:{f}_{{x}} \left(\mathrm{0},\mathrm{1}\right),{f}_{{y}} \left(\mathrm{2},−\mathrm{1}\right),\:{f}_{{xx}} \left(\mathrm{0},\mathrm{1}\right),{f}_{{xy}} \left(\mathrm{2},−\mathrm{1}\right) \\ $$ Commented…

solve-the-differential-equation-y-x-x-y-2-x-please-i-need-help-

Question Number 125901 by pticantor last updated on 15/Dec/20 $$\:\bigstar^{\bigstar\bigstar} \boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{differential}}\:\boldsymbol{{equation}}\bigstar^{\bigstar^{\bigstar} } \\ $$$$ \\ $$$$\:\:\boldsymbol{{y}}^{'} \left(\boldsymbol{{x}}\right)+\boldsymbol{{x}}=\boldsymbol{{y}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right) \\ $$$$ \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{i}}\:\boldsymbol{{need}}\:\boldsymbol{{help}}\:!! \\ $$ Terms…

Question-60357

Question Number 60357 by Sardor2211 last updated on 20/May/19 Commented by Mr X pcx last updated on 20/May/19 $${the}\:{equation}\:{is}\:{not}\:{clear}\:{but}\:{i}\:{suppose} \\ $$$${that}\:{is}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} −\mathrm{2}{xy}\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\…

Let-n-1-and-integer-P-n-X-1-X-n-1-X-n-1-Factorize-P-n-2-Deduce-S-n-k-1-n-4-cotan-2-kpi-2n-1-

Question Number 125760 by snipers237 last updated on 13/Dec/20 $${Let}\:{n}\geqslant\mathrm{1}\:{and}\:{integer},\:{P}_{{n}} \left({X}\right)=\left(\mathrm{1}+{X}\right)^{{n}} −\left(\mathrm{1}−{X}\right)^{{n}} \: \\ $$$$\left.\mathrm{1}\right)\:{Factorize}\:{P}_{{n}} \\ $$$$\left.\mathrm{2}\right){Deduce}\:\:{S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left[\mathrm{4}+{cotan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:\right] \\ $$ Answered by…