Question Number 53890 by shaddie last updated on 27/Jan/19 $$\mathrm{pls}.\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{this} \\ $$$$\mathrm{if}\:\mathrm{y}=\frac{\mathrm{x}−\mathrm{2}}{\mathrm{2x}−\mathrm{1}}\:\mathrm{show}\:\mathrm{that}\:\left(\mathrm{2x}−\mathrm{1}\right)\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\mathrm{4}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{0} \\ $$ Answered by math1967 last updated on 27/Jan/19 $$\Rightarrow{y}\left(\mathrm{2}{x}−\mathrm{1}\right)=\left({x}−\mathrm{2}\right) \\…
Question Number 53769 by ajfour last updated on 25/Jan/19 $${Find}\:{maximum}\:{of} \\ $$$$\:\:\:\frac{{xyz}}{\left({x}+{a}\right)\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{b}\right)}\:\:. \\ $$ Commented by mr W last updated on 25/Jan/19 $$\frac{\mathrm{1}}{\mathrm{16}\sqrt{{ab}}}\:? \\ $$…
Question Number 119280 by snipers237 last updated on 23/Oct/20 $$\:{Please}\:{more}\:{informations}\:{about}\:{this}\:{operator} \\ $$$$\:\:\underset{{p}=\mathrm{1}} {\overset{\infty} {{K}}}\left(……\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 184819 by cortano1 last updated on 12/Jan/23 $$\:\:{For}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:,\:{maximum}\:{value} \\ $$$$\:\:{of}\:{f}\left({x}\right)={x}\sqrt{\mathrm{1}−{x}+\sqrt{\mathrm{1}−{x}}}\:{is}\:\_\_ \\ $$ Answered by Frix last updated on 12/Jan/23 $${f}'\left({x}\right)=\mathrm{0} \\ $$$$−\frac{\mathrm{5}{x}−\mathrm{4}+\mathrm{2}\left(\mathrm{3}{x}−\mathrm{2}\right)\sqrt{\mathrm{1}−{x}}}{\mathrm{4}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{1}−{x}+\sqrt{\mathrm{1}−{x}}}}=\mathrm{0} \\…
Question Number 119240 by benjo_mathlover last updated on 23/Oct/20 $${find}\:{max}\:{and}\:{min}\:{value}\:{of}\: \\ $$$${f}\left({x},{y}\right)\:=\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{xy}+\mathrm{9}{y}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{24}{y}+\mathrm{4}\: \\ $$ Answered by 1549442205PVT last updated on 23/Oct/20 $${f}\left({x},{y}\right)\:=\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{xy}+\mathrm{9}{y}^{\mathrm{2}}…
Question Number 53676 by peter frank last updated on 24/Jan/19 $${If}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}\right) \\ $$$${show}\:{that} \\ $$$$\frac{{dy}}{{dx}}+\sqrt{\frac{{y}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}}\:=\mathrm{0} \\ $$$$ \\ $$ Commented by…
Question Number 184735 by mustafazaheen last updated on 11/Jan/23 $$ \\ $$$${f}\left({x},{y}\right)=\left(\sqrt{\mathrm{3}{xy}^{\mathrm{2}} }\right)\left(\sqrt[{\mathrm{5}}]{{x}^{\mathrm{5}} {y}^{\mathrm{2}} }\right) \\ $$$${f}^{'} \left({x},{y}\right)=?\:\:\:\:\:{f}''\left({x},{y}\right)=? \\ $$$$ \\ $$ Commented by mr…
Question Number 184720 by mustafazaheen last updated on 10/Jan/23 $$ \\ $$$${y}=\left(\sqrt{\mathrm{3}{xy}^{\mathrm{2}} }\right)\left(\sqrt[{\mathrm{5}}]{{x}^{\mathrm{5}} {y}^{\mathrm{2}} }\right) \\ $$$${y}'=?\:\:\:\:\:\:{y}^{''} =? \\ $$ Answered by Frix last updated…
Question Number 118969 by benjo_mathlover last updated on 21/Oct/20 $$\:{Given}\:{f}\left({x}\right)\:=\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}} \\ $$$${find}\:{the}\:{value}\:{of}\: \\ $$$$\:{f}\:''\left({x}\right)\:+\:{f}\:'\left({x}\right)\:+\:\mathrm{1}\:. \\ $$ Answered by TANMAY PANACEA last updated on 21/Oct/20 $${f}\left({x}\right)=\frac{{tanx}+\mathrm{1}}{{tanx}−\mathrm{1}}=−{tan}\left(\frac{\pi}{\mathrm{4}}+{x}\right)…
Question Number 53426 by Necxx last updated on 21/Jan/19 $${The}\:{general}\:{solution}\:{of}\:{the}\:{equation} \\ $$$$\frac{{dy}}{{dx}}+{ylnx}={x}^{−{x}} \\ $$$$\left.{a}\left.\right){x}^{{x}} \left(\mathrm{1}−{ce}^{{x}} \right)\:\:{b}\right)−{x}^{−{x}} \left(\mathrm{1}+{ce}^{\mathrm{2}{x}} \right) \\ $$$$\left.{c}\right)−{x}^{−{x}} \left(\mathrm{1}−{ce}^{{x}} \right) \\ $$ Commented…