Question Number 76714 by peter frank last updated on 29/Dec/19 $${If}\:{y}=\sqrt{\mathrm{tan}\:{x}+\sqrt{\mathrm{tan}\:{x}+\sqrt{\mathrm{tan}\:{x}+….\infty}}}\: \\ $$$${prove}\:{that} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{sec}\:^{\mathrm{2}} {x}}{\mathrm{2}{y}−\mathrm{1}} \\ $$ Answered by mind is power last updated…
Question Number 142248 by gsk2684 last updated on 28/May/21 Commented by gsk2684 last updated on 31/May/21 $$\mathrm{let}\:\mathrm{a}>\mathrm{0}\:\mathrm{and} \\ $$$$\mathrm{D}=\left\{\left(\mathrm{x},\mathrm{y}\right)/\mathrm{0}\leq\mathrm{x}\leq\mathrm{1},\mathrm{0}\leq\mathrm{y}\leq\mathrm{3x}\right\} \\ $$$$\mathrm{consider}\:\mathrm{the}\:\mathrm{following}\:\mathrm{integral} \\ $$$$\int\underset{\mathrm{D}} {\int}\left(\mathrm{3x}+\mathrm{y}\right)\left(\mathrm{3x}−\mathrm{y}\right)^{\mathrm{a}} \mathrm{dxdy}…
Question Number 76715 by peter frank last updated on 29/Dec/19 $${If}\:{u}={arc}\mathrm{sin}\:\frac{{x}}{{y}}+{arc}\mathrm{tan}\:\frac{{y}}{{x}} \\ $$$${show}\:{that}\: \\ $$$${x}\frac{\partial{u}}{{dx}}+{y}\frac{\partial{u}}{{dy}}=\mathrm{0} \\ $$ Commented by abdomathmax last updated on 30/Dec/19 $$\frac{\partial{u}}{\partial{x}}\:=\frac{\mathrm{1}}{{y}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}}…
Question Number 76713 by peter frank last updated on 29/Dec/19 $${If}\:\mathrm{cos}\:{y}={x}\mathrm{cos}\:\left({a}+{y}\right),{show} \\ $$$${that}\:\frac{{dy}}{{dx}}=\frac{\mathrm{cos}\:^{\mathrm{2}} \left({a}+{y}\right)}{\mathrm{sin}\:{a}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 142245 by mathmax by abdo last updated on 28/May/21 $$\mathrm{developp}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\lambda\:+\mathrm{sinx}}\left(\:\lambda\:\mathrm{real}\:>\mathrm{0}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76711 by aliesam last updated on 29/Dec/19 Answered by MJS last updated on 30/Dec/19 $${x}>\mathrm{0}\wedge{y}>\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} >\mathrm{0}\wedge{x}^{\mathrm{2}} +{y}^{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{0}<{x}^{\mathrm{3}} +{y}^{\mathrm{3}} \leqslant{x}−{y}\:\Rightarrow\:\mathrm{0}<{x}−{y}\:\Leftrightarrow\:{y}<{x}…
Question Number 142246 by mathdanisur last updated on 28/May/21 Answered by mr W last updated on 28/May/21 $${V}=\frac{\mathrm{1}}{\mathrm{6}}\begin{pmatrix}{{a}}&{{a}}&{{a}}\\{{b}}&{\mathrm{1}}&{{b}}\\{{a}}&{{b}}&{{b}}\end{pmatrix} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[{a}\left({b}−{b}^{\mathrm{2}} \right)−{a}\left({b}^{\mathrm{2}} −{ab}\right)+{a}\left({b}^{\mathrm{2}} −{a}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}{a}\left({b}−{a}\right)\left(\mathrm{1}−{b}\right)…
Question Number 11172 by agni5 last updated on 15/Mar/17 $$\mathrm{Tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} =\left(\mathrm{x}−\mathrm{y}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{at}\:\left(−\mathrm{1},\mathrm{1}\right). \\ $$ Answered by ajfour last updated on 15/Mar/17 $${y}={x}+\mathrm{2} \\ $$…
Question Number 11167 by agni5 last updated on 15/Mar/17 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ln}\mid\mathrm{x}+\sqrt{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mid\:\mathrm{is}\:\mathrm{even}\:\mathrm{or}\:\mathrm{odd}?\: \\ $$$$\mathrm{give}\:\mathrm{reasion}. \\ $$ Answered by mrW1 last updated on 15/Mar/17 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ln}\mid\mathrm{x}+\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\mid=\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)…
Question Number 142236 by mathocean1 last updated on 28/May/21 $$\mathrm{f}\:\mathrm{is}\:\mathrm{an}\:\mathrm{endomorphism}\:\mathrm{of}\:\mathrm{V}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{f}\circ\mathrm{f}=−\mathrm{Id}_{\mathrm{V}} \:. \\ $$$$\mathrm{1}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{f}\:\mathrm{is}\:\mathrm{an}\:\mathrm{isomorphism}\:\mathrm{of} \\ $$$$\mathrm{V}\:\mathrm{and}\:\mathrm{express}\:\mathrm{f}^{−\mathrm{1}} \:\mathrm{in}\:\mathrm{function}\:\mathrm{of}\:\mathrm{f}. \\ $$$$\mathrm{2}.\:\mathrm{show}\:\mathrm{that}\:\overset{\rightarrow} {\mathrm{0}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{one}\:\mathrm{invariant} \\ $$$$\mathrm{vector}\:\mathrm{by}\:\mathrm{f}. \\ $$$$\mathrm{3}.\:\mathrm{Given}\:\overset{\rightarrow}…