Question Number 39469 by Raj Singh last updated on 06/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18 $${let}\:{point}\:{be}\:\alpha\:,\beta \\ $$$${calculation}\:{of}\:{slope}\:{is}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{{y}}}×\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=−\frac{\sqrt{{y}}\:}{\:\sqrt{{x}}} \\…
Question Number 39466 by Raj Singh last updated on 06/Jul/18 Commented by MJS last updated on 06/Jul/18 $${y}={x}^{\mathrm{3}} −{x}^{\mathrm{2}} \:\mathrm{has}\:\mathrm{no}\:\mathrm{vertical}\:\mathrm{tangent} \\ $$ Terms of Service…
Question Number 39457 by Raj Singh last updated on 06/Jul/18 Answered by MrW3 last updated on 06/Jul/18 $${y}={f}\left({x}\right)={b}\:{e}^{−\frac{{x}}{{a}}} \\ $$$${f}\left(\mathrm{0}\right)={b} \\ $$$${y}'={f}'\left({x}\right)=−\frac{{b}}{{a}}\:{e}^{−\frac{{x}}{{a}}} \\ $$$${f}'\left(\mathrm{0}\right)=−\frac{{b}}{{a}} \\…
Question Number 104939 by ~blr237~ last updated on 24/Jul/20 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)!\left(\mathrm{2}{n}+\mathrm{1}\right)}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 104927 by bobhans last updated on 24/Jul/20 $$\int\frac{{dx}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} } \\ $$ Answered by Dwaipayan Shikari last updated on 24/Jul/20 $$\int\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{C}…
Question Number 104860 by ~blr237~ last updated on 24/Jul/20 $$\left(\mathrm{1}−{aT}\right)\left(\mathrm{1}−{bT}\right)…..\left(\mathrm{1}−{yT}\right)\left(\mathrm{1}−{zT}\right)\:\:\:\:???? \\ $$ Answered by ~blr237~ last updated on 24/Jul/20 $${Just}\:\:{to}\:{smile}\::\:{if}\:{we}\:{remember}\:{the}\:{formula}\:{of}\:{the}\:{frequency}\:\:\:{f}=\frac{\mathrm{1}}{{T}}\:\:{then}\:\:\mathrm{1}−{fT}=\mathrm{0}\: \\ $$$${And}\:{that}\:{product}\:{will}\:{be}\:{null} \\ $$ Commented…
Question Number 104856 by ~blr237~ last updated on 24/Jul/20 $$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\overset{−} {{z}}}{{z}}\:\:\:\:,\:\:\:\:\underset{{z}\rightarrow{i}} {\mathrm{lim}}\:\frac{\left(\overset{−} {{z}}\right)^{\mathrm{4}} }{{z}^{\mathrm{4}} }\:\:,\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sinz}}{{z}}\: \\ $$ Answered by abdomathmax last updated on…
Question Number 104858 by ~blr237~ last updated on 24/Jul/20 $$\:\:\left(\mathrm{1}−{a}^{\mathrm{3}} \right)\left(\mathrm{1}−{b}^{\mathrm{3}} \right)……\left(\mathrm{1}−{x}^{\mathrm{3}} \right)\left(\mathrm{1}−{y}^{\mathrm{3}} \right)\left(\mathrm{1}−{z}^{\mathrm{3}} \right)\:\:\:\:\:\:\:\:???? \\ $$ Commented by 1549442205PVT last updated on 24/Jul/20 $$\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{your}}\:\:\boldsymbol{\mathrm{question}}?…
Question Number 104857 by ~blr237~ last updated on 24/Jul/20 $$\:\:{y}'=\frac{{y}−{x}}{{y}+{x}}\:\:\:\:\:\:\:{y}\left(\mathrm{0}\right)=\mathrm{0} \\ $$ Answered by bemath last updated on 24/Jul/20 $${y}={px}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:{p}+{x}\frac{{dp}}{{dx}} \\ $$$$\Leftrightarrow{p}+{x}\frac{{dp}}{{dx}}\:=\:\frac{{x}\left({p}−\mathrm{1}\right)}{{x}\left({p}+\mathrm{1}\right)} \\ $$$${x}\frac{{dp}}{{dx}}\:=\:\frac{{p}−\mathrm{1}}{{p}+\mathrm{1}}−\frac{{p}\left({p}+\mathrm{1}\right)}{{p}+\mathrm{1}} \\…
Question Number 104852 by ~blr237~ last updated on 24/Jul/20 $$\:\:{if}\:\:{g}\in{C}\left(\mathbb{R},\mathbb{R}\right)\:{and}.\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{g}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{3}}+\int_{\mathrm{0}} ^{\mathrm{1}} {g}^{\mathrm{2}} \left({x}^{\mathrm{2}} \right){dx}\: \\ $$$${then}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {g}\left({x}\right){dx}=\frac{\mathrm{2}}{\mathrm{3}}\:{and}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {g}^{\mathrm{2}} \left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\:\: \\ $$…