Question Number 222329 by klipto last updated on 22/Jun/25 $$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\infty} \:\mathrm{4}\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{\mathrm{x}}} \\ $$$$ \\ $$ Commented by mr W last updated on 22/Jun/25 $$=\infty…
Question Number 222284 by klipto last updated on 21/Jun/25 $$\boldsymbol{\mathrm{y}}=\frac{\mathrm{8}^{\boldsymbol{\mathrm{x}}} }{\left(\boldsymbol{\mathrm{in}}\mathrm{8}\right)^{\mathrm{3}} } \\ $$$$\boldsymbol{\mathrm{find}}\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{6}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{6}} } \\ $$ Answered by mr W last updated on…
Question Number 222113 by klipto last updated on 17/Jun/25 $$\mathrm{name}\:\mathrm{the}\:\mathrm{following}\:\mathrm{compound} \\ $$ Commented by klipto last updated on 18/Jun/25 Commented by MathematicalUser2357 last updated on…
Question Number 221044 by fantastic last updated on 23/May/25 $${If}\:{V}\:{be}\:{a}\:{function}\:{of}\:{x}\:{and}\:{y},\:{prove}\:{that} \\ $$$$\frac{\partial^{\mathrm{2}} {V}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {V}}{\partial{y}^{\mathrm{2}} }=\frac{\partial^{\mathrm{2}} {V}}{\partial{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}}\:\frac{\partial{V}}{\partial{r}}+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{\partial^{\mathrm{2}} {V}}{\partial\theta^{\mathrm{2}} }, \\ $$$${where}\:{x}={r}\:\mathrm{cos}\:\theta\:,\:{y}={r}\mathrm{sin}\:\theta \\ $$…
Question Number 220964 by fantastic last updated on 21/May/25 $${Find}\:{the}\:{general}\:{solution}\:{of}\:{the}\:{differential}\:{equation} \\ $$$${x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:+\:{x}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{6}\frac{{dy}}{{dx}}+\mathrm{6}\frac{{y}}{{x}}=\frac{{x}\:\mathrm{ln}\:{x}+\mathrm{1}}{{x}^{\mathrm{2}} },\left[{x}>\mathrm{0}\right] \\ $$ Terms of Service Privacy Policy…
Question Number 220863 by fantastic last updated on 20/May/25 $$\left(\mathrm{211}\right) \\ $$$$\:\: \\ $$$${Find}\:{the}\:{derivative}\:{of}\:\Delta{x},\:{where} \\ $$$$\Delta{x}=\begin{vmatrix}{{f}_{\mathrm{1}} \left({x}\right)}&{\phi_{\mathrm{1}} \left({x}\right)}&{\Psi_{\mathrm{1}} \left({x}\right)}\\{{f}_{\mathrm{2}} \left({x}\right)}&{\phi_{\mathrm{2}} \left({x}\right)}&{\Psi_{\mathrm{2}} \left({x}\right)}\\{{f}_{\mathrm{3}} \left({x}\right)}&{\phi_{\mathrm{3}} \left({x}\right)}&{\Psi_{\mathrm{3}} \left({x}\right)}\end{vmatrix}…
Question Number 220232 by mnjuly1970 last updated on 09/May/25 $$ \\ $$$$\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\:\:\frac{\pi}{\mathrm{16}}\:<\:\int_{\mathrm{0}} ^{\:\mathrm{1}\:} \sqrt{\frac{{x}\left(\mathrm{1}−{x}\right)}{{sin}\left(\pi{x}\right)+{cos}\left(\pi{x}\right)+\mathrm{2}}}\:{dx}<\frac{\pi}{\mathrm{8}} \\ $$$$\:\:\:\:\: \\ $$ Answered by SdC355…
Question Number 220131 by fantastic last updated on 06/May/25 $${If}\:\:\:{f}\left({x},{y}\right)=\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{{n}} }{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}+{x}\phi\left(\frac{{y}}{{x}}\right)+\Psi\left(\frac{{y}}{{x}}\right), \\ $$$${then}\:{using}\:{Euler}'{s}\:{theorem}\:{on}\:{homogenous}\:{functions},{show}\:{that} \\ $$$${x}^{\mathrm{2}} \frac{\delta^{\mathrm{2}} {f}}{\delta{x}^{\mathrm{2}} }+\mathrm{2}{xy}\frac{\delta^{\mathrm{2}} {f}}{\delta{x}\delta{y}}+{y}^{\mathrm{2}} \frac{\delta^{\mathrm{2}} {f}}{\delta{y}^{\mathrm{2}} }=\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}}…
Question Number 219868 by Nicholas666 last updated on 02/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\frac{{d}}{{dx}}\:\left(\frac{\mathrm{sin}^{\:\mathrm{2}} {x}}{\mathrm{1}+\mathrm{cot}\:{x}}\:+\:\frac{\mathrm{cos}^{\:\mathrm{2}} {x}}{\mathrm{1}+\mathrm{tan}\:{x}}\right)\:=\:−\mathrm{cos}\:\mathrm{2}{x}\:\:\:\: \\ $$$$ \\ $$ Answered by MrGaster last updated…
Question Number 219451 by Nicholas666 last updated on 25/Apr/25 Commented by Nicholas666 last updated on 25/Apr/25 $$\mathrm{This}\:\mathrm{is}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{beyond}\:\mathrm{My}\:\mathrm{control},\:\:\: \\ $$$$\:\mathrm{can}\:\mathrm{You}\:\mathrm{solve}\:\mathrm{friends}? \\ $$$$ \\ $$ Terms of…