Question Number 208553 by efronzo1 last updated on 18/Jun/24 $$\:\:\:\cancel{ } \\ $$ Answered by mr W last updated on 18/Jun/24 $${due}\:{to}\:{symmetry} \\ $$$${y}={x}\:\Rightarrow\mathrm{3}{x}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{x}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}…
Question Number 208499 by efronzo1 last updated on 17/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 207954 by solihin last updated on 01/Jun/24 $$\:\underline{\boldsymbol{{x}}} \\ $$ Commented by MathematicalUser2357 last updated on 02/Jun/24 $${I}\:{see}\:\underline{\boldsymbol{{x}}}\:{in}\:{tinku}\:{tara} \\ $$ Answered by AliJumaa…
Question Number 207175 by MATHEMATICSAM last updated on 08/May/24 $$\mathrm{If}\:{x}^{{m}} .{y}^{{n}} \:=\:\left({x}\:+\:{y}\right)^{{m}\:+\:{n}} \:\mathrm{then}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:? \\ $$ Answered by som(math1967) last updated on 08/May/24 $${lnx}^{{m}}…
Question Number 207122 by MATHEMATICSAM last updated on 07/May/24 $$\mathrm{If}\:{e}^{{y}} \left({x}\:+\:\mathrm{1}\right)\:=\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} . \\ $$ Answered by BaliramKumar last updated on 07/May/24…
Question Number 207109 by MATHEMATICSAM last updated on 06/May/24 $$\mathrm{If}\:{y}\:=\:\left(\mathrm{1}\:+\:{x}\right)\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\left(\mathrm{1}\:+\:{x}^{\mathrm{4}} \right)\:….\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}{n}} \right) \\ $$$$\mathrm{then}\:\mathrm{find}\:\frac{{dy}}{{dx}}\:\mathrm{at}\:{x}\:=\:\mathrm{0}. \\ $$ Answered by Berbere last updated on 06/May/24 $${y}\left({x}\right)=\left(\mathrm{1}+{x}\right)\underset{{k}=\mathrm{1}}…
Question Number 206882 by efronzo1 last updated on 29/Apr/24 $$\:\:{f}\left({x}\right)=\mathrm{tan}\:^{\mathrm{2}} {x}\:\sqrt{\mathrm{tan}\:{x}\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}\sqrt[{\mathrm{5}}]{\mathrm{tan}\:{x}\sqrt{…}}}}} \\ $$$$\:{f}\:'\left(\frac{\pi}{\mathrm{4}}\right)=? \\ $$ Answered by MM42 last updated on 29/Apr/24 $${f}\left({x}\right)={tan}^{\mathrm{2}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{6}}}…
Question Number 206340 by mnjuly1970 last updated on 12/Apr/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\:{ln}\left(\mathrm{1}−{x}\:\right){ln}\left(\mathrm{1}+{x}\:\right)}{{x}}{dx}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\Omega_{{n}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:{find}\::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:{n}\:\Omega_{{n}} \:=\:? \\ $$…
Question Number 206244 by cortano21 last updated on 10/Apr/24 $$\:\:\:\zeta \\ $$ Answered by A5T last updated on 10/Apr/24 $$\pi{r}_{{a}} ^{\mathrm{3}} =\pi{r}_{{b}} ^{\mathrm{3}} ×\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\frac{{r}_{{a}} }{{r}_{{b}}…
Question Number 205827 by mustafazaheen last updated on 31/Mar/24 $$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{implceat}\:\mathrm{second}\:\mathrm{derivative} \\ $$ Answered by cortano12 last updated on 31/Mar/24 $$\:\mathrm{3}{x}^{\mathrm{2}} +\:\mathrm{3}{y}^{\mathrm{2}}…