Question Number 134285 by bramlexs22 last updated on 02/Mar/21 $$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\left(\mathrm{16}−\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)}\: \\ $$ Answered by EDWIN88 last updated on 02/Mar/21 $$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)=\left(\mathrm{16}−\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}}…
Question Number 68703 by Maclaurin Stickker last updated on 15/Sep/19 $$\frac{{d}}{{dx}}\left({ln}\left(\sqrt{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}}\right)\right)=? \\ $$ Commented by MJS last updated on 15/Sep/19 $$\frac{{d}}{{dx}}\left[\mathrm{ln}\:\sqrt{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}}\right]=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{d}}{{dx}}\left[\mathrm{ln}\:\frac{{x}^{\mathrm{2}}…
Question Number 134204 by liberty last updated on 01/Mar/21 $$\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2x}\right)}\:. \\ $$$$\mathrm{find}\:\mathrm{f}\:'\left(\mathrm{0}\right). \\ $$ Answered by bramlexs22 last updated on 02/Mar/21 $$\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{2}\left(\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}\right)}\:=\mid\mathrm{4sin}\:\mathrm{x}\:\mid=\mathrm{2}\mid\mathrm{sin}\:\mathrm{x}\:\mid \\ $$$$\:\mathrm{f}\:'\left(\mathrm{0}^{−}…
Question Number 134127 by bobhans last updated on 28/Feb/21 $$\:\mathrm{Given}\:\mid\mathrm{f}\left(\mathrm{x}\right)\mid\:=\:\mathrm{x}^{\mathrm{2}} \:\mathrm{for}\:−\mathrm{1}<\mathrm{x}<\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{f}\:'\left(\mathrm{0}\right). \\ $$ Commented by EDWIN88 last updated on 28/Feb/21 $$\mathrm{does}\:\mathrm{not}\:\mathrm{exist} \\ $$…
Question Number 134108 by Eric002 last updated on 27/Feb/21 $$\frac{{d}}{{dx}}\left(\sqrt[{\mathrm{4}}]{\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}}\right) \\ $$ Answered by Ñï= last updated on 27/Feb/21 $$\frac{{d}}{{dx}}\left(\sqrt[{\mathrm{4}}]{\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}}\right) \\…
Question Number 134102 by mnjuly1970 last updated on 27/Feb/21 $$\:\:\:\:\:\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\sqrt{{x}}\:\right)}{{e}^{\mathrm{2}\pi\sqrt{{x}}\:} −\mathrm{1}}{dx}=\mathrm{1}−\frac{{e}}{\left({e}−\mathrm{1}\right)_{} ^{\mathrm{2}} } \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 134040 by benjo_mathlover last updated on 27/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance}\: \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$$\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}−\mathrm{1and}\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}−\mathrm{1} \\ $$ Commented by benjo_mathlover last updated on 27/Feb/21…
Question Number 133994 by benjo_mathlover last updated on 26/Feb/21 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{function}\:\mathrm{f}\:\mathrm{satisfies} \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{{x}+{a}\sqrt{\mathrm{2}}\:\mathrm{sin}\:{x}\:;\:\mathrm{0}\leqslant{x}<\frac{\pi}{\mathrm{4}}}\\{\mathrm{2}{x}\:\mathrm{cot}\:{x}\:+{b}\:;\:\frac{\pi}{\mathrm{4}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}}\\{{a}\:\mathrm{cos}\:\mathrm{2}{x}−{b}\mathrm{sin}\:{x}\:;\:\frac{\pi}{\mathrm{2}}<{x}\leqslant\pi}\end{cases} \\ $$$$\:\mathrm{continuous}\:\mathrm{in}\:\left[\:\mathrm{0},\pi\:\right],\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{and}\:{b}.\: \\ $$ Answered by EDWIN88 last updated on 26/Feb/21…
Question Number 133928 by mnjuly1970 last updated on 25/Feb/21 Answered by mathmax by abdo last updated on 25/Feb/21 $$\left.\mathrm{2}\right)\:\mathrm{u}_{\mathrm{n}} =\left(\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \:=\frac{\left(\mathrm{2n}\right)!}{\left(\mathrm{n}!\right)^{\mathrm{2}}…
Question Number 133915 by nherly last updated on 25/Feb/21 Commented by TheSupreme last updated on 25/Feb/21 $${nice}\:{photo} \\ $$ Terms of Service Privacy Policy Contact:…