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)} \\ $$$${we}\:{want}\:{to}\:{compute}\:{the}\:{value}\:{of}\: \\ $$$$\left({f}^{−\mathrm{1}} \right)'\left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{{f}\:'\left({f}^{−\mathrm{1}} \left(\mathrm{2}\right)\right)}…
Question Number 123410 by bemath last updated on 25/Nov/20 $$\:{Suppose}\:{that}\:{f}\:{is}\:{differentiable} \\ $$$${with}\:{derivative}\:{f}\:'\left({x}\right)=\:\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{−\mathrm{1}/\mathrm{2}} . \\ $$$${Show}\:{that}\:{g}\:=\:{f}^{−\mathrm{1}} \:{satisfies}\: \\ $$$${g}''\left({x}\right)=\:\frac{\mathrm{3}}{\mathrm{2}}{g}\left({x}\right)^{\mathrm{2}} \\ $$ Answered by mnjuly1970 last…
Question Number 123407 by bemath last updated on 25/Nov/20 $${Find}\:{the}\:{all}\:{asymtotes}\:{of}\:{the} \\ $$$${curve}\:{y}\:=\:\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}{{x}−\mathrm{2}} \\ $$ Commented by EVIMENEBASSEY last updated on 25/Nov/20 $$ \\ $$$$…
Question Number 123396 by bemath last updated on 25/Nov/20 $$\:{Find}\:{the}\:{greatest}\:{and}\:{the}\:{least}\:{values} \\ $$$${of}\:{the}\:{function}\:{on}\:{indicates} \\ $$$${interval}\:\left(−\infty,\infty\right)\: \\ $$$$\left({i}\right)\:{y}\:=\:\mathrm{sin}\:{x}\:\mathrm{sin}\:\mathrm{2}{x}\:. \\ $$ Answered by TANMAY PANACEA last updated on…
Question Number 57820 by maxmathsup by imad last updated on 12/Apr/19 $${solve}\:\sqrt{{x}+\mathrm{1}}{y}^{'} −\sqrt{{x}−\mathrm{2}}{y}\:={x}^{\mathrm{2}} \:{e}^{−\mathrm{2}{x}} \:\:\:{with}\:{y}\left(\mathrm{3}\right)\:=\mathrm{1} \\ $$ Commented by maxmathsup by imad last updated on…