Question Number 145620 by Study last updated on 06/Jul/21 $$\frac{{d}}{{dx}}\left(\sqrt[{\mathrm{3}}]{{x}+\sqrt[{\mathrm{3}}]{{x}+\sqrt[{\mathrm{3}}]{{x}+…}}}\right)=? \\ $$ Answered by Olaf_Thorendsen last updated on 06/Jul/21 $${y}\:=\:{f}\left({x}\right)\:=\:\sqrt[{\mathrm{3}}]{{x}+{f}\left({x}\right)}\:=\:\sqrt[{\mathrm{3}}]{{x}+{y}} \\ $$$${y}^{\mathrm{3}} \:=\:{x}+{y} \\ $$$$\mathrm{3}{y}^{\mathrm{2}}…
Question Number 145602 by mnjuly1970 last updated on 06/Jul/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\mathrm{Advanced}\:………\mathrm{Calculus}….. \\ $$$$\:\:\:\mathrm{Q}::\:\:\:\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|c|}{\:{i}\:::\:\:\:\boldsymbol{\phi}\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{Ln}\:\left(\:\Gamma\:\left(\:\mathrm{2}\:+\:{x}\:\right)\:\right){dx}\:=\:?\:\:\:\:}\\{\:{ii}\:::\:\:\:\Omega\::=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{1}}{\:{n}\:\left(\:\mathrm{2}{n}\:+\:\mathrm{3}\:\right)}\:=\:?}\\\hline\end{array} \\ $$$$ \\…
Question Number 14513 by tawa tawa last updated on 01/Jun/17 $$\mathrm{If}\:\:\mathrm{y}^{\mathrm{2}} \left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)\:=\:\mathrm{1}\:−\:\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\:\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{1}\:−\:\mathrm{y}^{\mathrm{4}} }{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{4}} } \\ $$ Answered by mrW1 last updated…
Question Number 80027 by jagoll last updated on 30/Jan/20 $${find}\:{minimum} \\ $$$${value}\:{of}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{153}} \\ $$$${for}\:{x}\geqslant\mathrm{0}\:{in}\:\mathbb{R} \\ $$ Commented by john santu last updated on…
Question Number 79964 by ubaydulla last updated on 29/Jan/20 $$\int\left(\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{4}} {x}\right){dx}=\left[\mathrm{tan}\:^{\mathrm{2}} {x}={t}\right]=\int\left({t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){dx}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{{t}^{\mathrm{5}} }{\mathrm{5}}+{c} \\ $$ Commented by $@ty@m123 last updated on…
Question Number 145449 by solihin last updated on 05/Jul/21 Answered by Olaf_Thorendsen last updated on 05/Jul/21 $${p}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\pi{a}+\mathrm{2}{a}+\mathrm{2}{b}\:=\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}\right){a}+\mathrm{2}{b}\:=\:\mathrm{5} \\ $$$${A}\:=\:\frac{\pi{a}^{\mathrm{2}} }{\mathrm{8}}+{ab} \\ $$$${A}\:=\:\frac{\pi{a}^{\mathrm{2}} }{\mathrm{8}}+{a}×\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}−\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}\right){a}\right) \\ $$$${A}\:=\:−\left(\frac{\pi}{\mathrm{8}}+\mathrm{1}\right){a}^{\mathrm{2}}…
Question Number 145450 by solihin last updated on 05/Jul/21 Answered by Olaf_Thorendsen last updated on 05/Jul/21 $$\left.{a}\right) \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{2}^{{n}} }{\mathrm{3}^{{n}+\mathrm{1}} },\:{n}\in\mathbb{N}^{\ast} \\ $$$${a}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}}…
Question Number 145399 by Willson last updated on 04/Jul/21 $$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow+\infty} {\boldsymbol{\mathrm{lim}}}\:\:\underset{\:\mathrm{0}} {\int}^{\:\boldsymbol{\mathrm{n}}} \:\frac{\boldsymbol{\mathrm{t}}^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}!}\:\boldsymbol{{e}}^{−\boldsymbol{\mathrm{t}}} \:\boldsymbol{\mathrm{dt}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by ArielVyny last updated on…
Question Number 79689 by ~blr237~ last updated on 27/Jan/20 $${Find}\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:−{x}\:\:\:\:\:{without}\:\:{DL}\:{and}\:{if}\:{you}\:{have}\:{to}\:{use}\:{the}\: \\ $$$${hospital}\:{rule}\:{please}\:\:{justify}\:{that}\:{your}\:{function}\:{is}\:{C}^{\mathrm{1}} \: \\ $$ Commented by mr W last updated on 27/Jan/20…
Question Number 79679 by ~blr237~ last updated on 27/Jan/20 $${Find}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+{x}\right)−{x} \\ $$ Commented by mathmax by abdo last updated on 27/Jan/20 $${let}\:{f}\left({x}\right)={x}^{\mathrm{2}} {ln}\left(\mathrm{1}+{x}\right)−{x}\:\Rightarrow{f}\left({x}\right)={x}^{\mathrm{2}}…