Question Number 352 by Vishal Bhardwaj last updated on 25/Jan/15 $${Q}.\:\:{Define}\:\bullet\bullet\:\:{LHD}\left({left}\:{hand}\right. \\ $$$$\left.{derivative}\right)\:=\:{f}\left({a}−\mathrm{0}\right) \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {{lim}}\:\frac{{f}\left({a}−{h}\right)−{f}\left({a}\right)}{−{h}} \\ $$$$ \\ $$$$ \\ $$ Commented by prakash…
Question Number 356 by 123456 last updated on 25/Jan/15 $${f}_{\mathrm{0}} \left({x}\right)=\mathrm{1} \\ $$$${f}_{\mathrm{1}} \left({x}\right)={x} \\ $$$${f}_{{n}+\mathrm{1}} \left({x}\right)={x}^{{f}_{{n}} \left({x}\right)} \\ $$$$\frac{\partial{f}_{{n}} }{\partial{x}}=?,{n}\in\mathbb{N}^{\ast} \\ $$ Answered by…
Question Number 295 by defg last updated on 25/Jan/15 $$\mathrm{If}\:{V}=\mathrm{log}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:\mathrm{then}\:{V}_{{xx}} +{V}_{{yy}} =? \\ $$ Answered by 123456 last updated on 19/Dec/14 $${V}_{{x}} =\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}}…
Question Number 281 by samarth last updated on 25/Jan/15 $$\mathrm{If}\:{f}\left({x}\right)=\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}},\:\mathrm{0}\leqslant{x}\leqslant\pi/\mathrm{2},\:\mathrm{then}\:{f}\:'\left(\pi/\mathrm{6}\right)=? \\ $$ Answered by 123456 last updated on 18/Dec/14 $${f}\left({x}\right)=\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}} \\ $$$$\frac{\partial{f}}{\partial{x}}=\frac{\partial}{\partial{x}}\left(\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}}\right)…
Question Number 131275 by Engr_Jidda last updated on 03/Feb/21 $${Orthogonal}\:{and}\:{Orthonormal}\:{sets} \\ $$$${given}\:{that}\:\:{y}_{{n}} \left({t}\right)=\varrho^{{t}} {B}_{{n}} {sin}\frac{{n}\pi}{\mathrm{4}}{t}\: \\ $$$${and}\:{y}_{{m}} \left({t}\right)=\varrho^{{t}} {B}_{{m}} {sin}\frac{{m}\pi}{\mathrm{4}}{t} \\ $$$${show}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{4}} {y}_{{n}} \left({t}\right){y}_{{m}}…
Question Number 65651 by Masumsiddiqui399@gmail.com last updated on 01/Aug/19 Commented by Prithwish sen last updated on 01/Aug/19 $$\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\mathrm{x}−\mathrm{1}=\mathrm{0}\:\mathrm{and}\:\mathrm{y}=\mathrm{0} \\ $$$$\therefore\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\mathrm{1} \\…
Question Number 131132 by 676597498 last updated on 01/Feb/21 $$\mathrm{u}\left(\mathrm{x}\right)=\mathrm{cosh}\left(\mathrm{x}\right)−\mathrm{x} \\ $$$$\mathrm{v}_{\mathrm{n}+\mathrm{1}} =\mathrm{u}\left(\mathrm{v}_{\mathrm{n}} \right) \\ $$$$\mathrm{m}=\sqrt{\mathrm{2}}−\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{v}_{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{A}.\:\mathrm{V}_{\mathrm{n}} −\mathrm{V}_{\mathrm{0}} =\mathrm{nm} \\ $$$$\mathrm{B}.\:\mathrm{v}_{\mathrm{n}}…
Question Number 131121 by abdurehime last updated on 01/Feb/21 Answered by mathmax by abdo last updated on 01/Feb/21 $$\left.\mathrm{A}\right)\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\mathrm{25}\:\Rightarrow\mathrm{y}^{\mathrm{2}} \:=\mathrm{25}−\mathrm{x}^{\mathrm{2}} \:\Rightarrow\mathrm{y}=\xi\sqrt{\mathrm{25}−\mathrm{x}^{\mathrm{2}} }\left(\:\xi=\overset{−} {+}\mathrm{1}\right)\:\Rightarrow…
Question Number 131104 by mnjuly1970 last updated on 01/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{b}−\mathrm{1}} }{\left(\mathrm{1}+{x}^{{a}} \right)^{{s}} }{dx}\overset{???} {=}\frac{\Gamma\left(\frac{{b}}{{a}}\right)\Gamma\left({s}−\frac{{b}}{{a}}\right)}{{a}\Gamma\left({s}\right)} \\ $$ Answered by Ar Brandon last…
Question Number 29 by user1 last updated on 25/Jan/15 $$\mathrm{Differentiate}\:\:\:{e}^{\sqrt{\mathrm{cot}\:{x}}} . \\ $$ Answered by user2 last updated on 03/Nov/14 $$\mathrm{Let}\:\:\:{y}={e}^{\sqrt{\mathrm{cot}\:{x}}} \\ $$$$\mathrm{Put}\:\mathrm{cot}\:{x}\:={t}\:\mathrm{and}\:\sqrt{\mathrm{cot}\:{x}}=\sqrt{{t}}={u},\:\mathrm{so}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}={e}^{{u}}…