Question Number 123763 by mnjuly1970 last updated on 28/Nov/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:{calculus}\:… \\ $$$$\:\:\:\:\:\mathscr{E}{valuate}\:::\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:=??? \\ $$ Answered by Snail last updated on 28/Nov/20 $$\underset{{n}=\mathrm{0}}…
Question Number 123752 by liberty last updated on 27/Nov/20 $${Let}\:{f}\left({x}\right)\:{be}\:{differentiable}\:{function} \\ $$$${and}\:{g}\left({x}\right)\:{be}\:{the}\:{inverse}\:{of}\:{f}\left({x}\right). \\ $$$${when}\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\left(\frac{{f}\left({x}\right)−{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} −\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:{then}\:\frac{{dg}\left({x}\right)}{{dx}}\mid_{{x}=\mathrm{8}} \:=? \\ $$ Commented by benjo_mathlover last updated…
Question Number 123720 by mnjuly1970 last updated on 27/Nov/20 $$\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:{calulus}… \\ $$$$\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\Phi=\:\:\int_{\mathrm{0}} ^{\:\:\infty} \frac{{x}^{\mathrm{3}} {e}^{\frac{−{x}}{\mathrm{2}}} }{{sinh}\left(\frac{{x}}{\mathrm{2}}\right)}\:=??? \\ $$ Answered by Dwaipayan Shikari last…
Question Number 123667 by sahnaz last updated on 27/Nov/20 $$\mathrm{y}=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\mathrm{d}^{\mathrm{2}} \mathrm{y}=? \\ $$ Answered by mathmax by abdo last updated on 27/Nov/20 $$\mathrm{y}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}}…
Question Number 58097 by rahul 19 last updated on 17/Apr/19 $${Find}\:{the}\:{angle}\:{between}\:{the}\:{curves}: \\ $$$$\left.\mathrm{1}\right){x}^{\mathrm{2}} {y}=\mathrm{1}−{y}\:{and}\:{x}^{\mathrm{3}} =\mathrm{2}−\mathrm{2}{y}. \\ $$$$\left.\mathrm{2}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \sqrt{\mathrm{2}}\:{and}\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={a}^{\mathrm{2}} . \\ $$…
Question Number 123575 by bramlexs22 last updated on 26/Nov/20 Commented by liberty last updated on 26/Nov/20 $$\left({i}\right)\:{grad}\:{normal}\:{a}\:{circle}\:{at}\:{point}\: \\ $$$${P}\left({a},\sqrt{\mathrm{4}−{a}^{\mathrm{2}} \:}\right)\:{is}\:{m}_{\mathrm{1}} =\:\frac{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}{{a}}\:{then}\: \\ $$$${eq}\:{of}\:{normal}\:{line}\:\Rightarrow{y}=\frac{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}{{a}}\left({x}−{a}\right)+\sqrt{\mathrm{4}−{a}^{\mathrm{2}}…
Question Number 58027 by problem solverd last updated on 16/Apr/19 Answered by tanmay last updated on 16/Apr/19 $$\mathrm{7}{xy}^{\mathrm{6}} \frac{{dy}}{{dx}}+{y}^{\mathrm{7}} −{x}^{\mathrm{5}} \frac{{dy}}{{dx}}−\mathrm{5}{x}^{\mathrm{4}} {y}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }=\mathrm{4}{y}+\mathrm{4}{x}\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}\left(\mathrm{7}{xy}^{\mathrm{6}}…
Question Number 123547 by mnjuly1970 last updated on 26/Nov/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{nice}\:\:\:\:\:{calculus}…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:\:{that}:::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\left\{\frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}\right\}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{3}}{log}^{\mathrm{3}} \left(\mathrm{2}\right)\checkmark \\ $$$$ \\ $$ Answered…
Question Number 123417 by sdfg last updated on 25/Nov/20 Answered by TANMAY PANACEA last updated on 25/Nov/20 $$\frac{{dy}}{{dx}}−\mathrm{2}{y}=\mathrm{3}{e}^{{x}} \\ $$$${e}^{−\mathrm{2}{x}} \frac{{dy}}{{dx}}−\mathrm{2}{ye}^{−\mathrm{2}{x}} =\mathrm{3}{e}^{−{x}} \:\:\left[{intregating}\:{factor}={e}^{\int{pdx}} ={e}^{\int−\mathrm{2}{dx}} \right.…
Question Number 123412 by bemath last updated on 25/Nov/20 Answered by liberty last updated on 25/Nov/20 $${use}\:{the}\:{formula}\:\left({f}^{−\mathrm{1}} \right)'\left({a}\right)\:=\:\frac{\mathrm{1}}{{f}\:'\left({f}^{−\mathrm{1}} \left({a}\right)\right)} \\ $$$${we}\:{want}\:{to}\:{compute}\:{the}\:{value}\:{of}\: \\ $$$$\left({f}^{−\mathrm{1}} \right)'\left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{{f}\:'\left({f}^{−\mathrm{1}} \left(\mathrm{2}\right)\right)}…