Question Number 127731 by Tosin Okunowo last updated on 01/Jan/21 $$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:……..\left[\mathrm{1}\right] \\ $$$$\mathrm{At}\:\mathrm{Q},\:\mathrm{y}\:+\:\delta\mathrm{y}\:=\:\left(\mathrm{x}+\delta\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{y}+\delta\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}.\delta\mathrm{x}+\left(\delta\mathrm{x}\right)^{\mathrm{2}} …..\left[\mathrm{2}\right] \\ $$$$\mathrm{Subtracting}\:\left[\mathrm{1}\right]\:\mathrm{from}\:\left[\mathrm{2}\right] \\ $$$$\mathrm{y}+\delta\mathrm{y}=\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}.\delta\mathrm{x}+\left(\delta\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{y}\:\:\:\:\:\:\:\:\:=\:\mathrm{x}^{\mathrm{2}}…
Question Number 127715 by snipers237 last updated on 01/Jan/21 $$\:\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:\frac{\left({argshx}\right)^{\mathrm{2}} }{\mathrm{2}{cosx}}\:{dx}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127649 by mnjuly1970 last updated on 31/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:\:\:{calculate}:: \\ $$$$\:\:\:\:\:\emptyset=\int_{\mathrm{0}} ^{\:\infty} {arctan}\left(\frac{\mathrm{1}}{{x}}\right).{sin}\left({x}\right){dx}=? \\ $$$$ \\ $$ Answered by mindispower last updated…
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Question Number 62102 by rajesh4661kumar@gamil.com last updated on 15/Jun/19 Answered by Kunal12588 last updated on 15/Jun/19 $$\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }+\sqrt{\mathrm{1}−{y}^{\mathrm{6}} }=\mathrm{9}\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\frac{−\mathrm{6}{x}^{\mathrm{5}} }{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }}+\frac{−\mathrm{6}{y}^{\mathrm{5}}…
Question Number 193065 by MikeH last updated on 03/Jun/23 Commented by MikeH last updated on 03/Jun/23 $$\mathrm{Find}\:{V}_{\mathrm{out}\:} \:\mathrm{from}\:\mathrm{the}\:\mathrm{circuit}\:\mathrm{above}.\: \\ $$ Answered by aleks041103 last updated…
Question Number 127454 by snipers237 last updated on 29/Dec/20 $$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{k}!\left({n}−{k}\right)!}{{n}!}\:=\:\mathrm{1} \\ $$ Answered by Ar Brandon last updated on 29/Dec/20 $$\forall\mathrm{n}\geqslant\mathrm{4},\:\forall\mathrm{k}\in\left\{\mathrm{2},…,\mathrm{n}−\mathrm{2}\right\},\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{k}}…
Question Number 127455 by snipers237 last updated on 29/Dec/20 $${Prove}\:{that}\:{for}\:{all}\:{n}\geqslant\mathrm{1}\: \\ $$$$\left.\mathrm{1}\left.\right){There}\:{exist}\:\:{a}_{{n}} \in\right]\mathrm{0},\mathrm{1}\left[\:{such}\:{as}\:\:\right. \\ $$$${sin}\left(\frac{\mathrm{1}}{{n}}\right)=\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{3}} }{cos}\left(\frac{\mathrm{1}}{{n}}{a}_{{n}} \right) \\ $$$$\left.\mathrm{2}\right)\:{Prove}\:{that}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{10}}\: \\ $$ Terms of…
Question Number 127452 by snipers237 last updated on 29/Dec/20 $$\:{Let}\:\:{a}\in\mathbb{R}^{\ast} ,\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({asin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{{a}}{cos}\left({n}\right)\right)^{{n}} =\mathrm{0} \\ $$ Answered by Ar Brandon last updated on 29/Dec/20 $$\mathrm{u}_{\mathrm{n}}…
Question Number 192901 by cortano12 last updated on 30/May/23 Answered by Frix last updated on 30/May/23 $$\mathrm{Assuming}\:{a},\:{b}\:>\mathrm{0} \\ $$$${z}=\frac{{x}}{{a}+\frac{{x}}{{b}+{z}}}\:\Rightarrow\:{z}=\frac{−{b}+\sqrt{\left(\mathrm{4}{x}+{ab}\right){b}}}{\mathrm{2}} \\ $$$${y}={x}+{z}\:\Rightarrow\:{y}={x}+\frac{−{b}+\sqrt{\left(\mathrm{4}{x}+{ab}\right){b}}}{\mathrm{2}} \\ $$$$\frac{{d}\left[{x}+\frac{−{b}+\sqrt{\left(\mathrm{4}{x}+{ab}\right){b}}}{\mathrm{2}}\right]}{{dx}}=\mathrm{1}+\frac{\sqrt{{b}}}{\:\sqrt{\left(\mathrm{4}{x}+{ab}\right){a}}} \\ $$…