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Category: Differentiation

Question-62102

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}}…

lim-n-k-1-n-k-n-k-n-1-

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}}…

Prove-that-for-all-n-1-1-There-exist-a-n-0-1-such-as-sin-1-n-1-n-1-6n-3-cos-1-n-a-n-2-Prove-that-lim-n-a-n-1-10-

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-192901

Question Number 192901 by cortano12 last updated on 30/May/23 Answered by Frix last updated on 30/May/23 $$\mathrm{Assuming}\:{a},\:{b}\:>\mathrm{0} \\ $$$${z}=\frac{{x}}{{a}+\frac{{x}}{{b}+{z}}}\:\Rightarrow\:{z}=\frac{−{b}+\sqrt{\left(\mathrm{4}{x}+{ab}\right){b}}}{\mathrm{2}} \\ $$$${y}={x}+{z}\:\Rightarrow\:{y}={x}+\frac{−{b}+\sqrt{\left(\mathrm{4}{x}+{ab}\right){b}}}{\mathrm{2}} \\ $$$$\frac{{d}\left[{x}+\frac{−{b}+\sqrt{\left(\mathrm{4}{x}+{ab}\right){b}}}{\mathrm{2}}\right]}{{dx}}=\mathrm{1}+\frac{\sqrt{{b}}}{\:\sqrt{\left(\mathrm{4}{x}+{ab}\right){a}}} \\ $$…