Question Number 204800 by sonukgindia last updated on 27/Feb/24 Answered by mr W last updated on 27/Feb/24 Commented by mr W last updated on 27/Feb/24…
Question Number 204715 by 073 last updated on 26/Feb/24 $${prove}\:{that} \\ $$$$\overset{\mathrm{1}} {\int}_{\mathrm{0}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$ Answered by witcher3 last updated on…
Question Number 204689 by Davidtim last updated on 25/Feb/24 $${y}=\mid{f}\left({x}\right)\mid\:\:;\:\:\:\:\frac{{dy}}{{dx}}=? \\ $$ Answered by A5T last updated on 25/Feb/24 $${y}=\sqrt{\left({f}\left({x}\right)\right)^{\mathrm{2}} }=\left[\left({f}\left({x}\right)\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\left({f}\left({x}\right)\right)^{\mathrm{2}} }}×\mathrm{2}{f}\left({x}\right)×{f}'\left({x}\right)=\frac{{f}\left({x}\right){f}^{'}…
Question Number 204688 by Davidtim last updated on 25/Feb/24 $${f}\left({x}\right)={sgn}\left({x}\right);\:\:\:\:\:{f}^{'} \left({x}\right)=\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]=? \\ $$ Answered by Faetmaaa last updated on 27/Feb/24 $$\frac{\mathrm{d}}{\mathrm{d}{x}}\left[{f}\mid_{\left.\right]−\infty,\:\mathrm{0}\left[\right.} \left({x}\right)\right]\:=\:\frac{\mathrm{d}}{\mathrm{d}{x}}\left[{f}\mid_{\left.\right]\mathrm{0},\:+\infty\left[\right.} \left({x}\right)\right]\:=\:\mathrm{0} \\ $$…
Question Number 204691 by Davidtim last updated on 25/Feb/24 $$\sqrt{{a}\boldsymbol{\div}\sqrt{{a}\boldsymbol{\div}\sqrt{{a}\boldsymbol{\div}\centerdot\centerdot\centerdot\boldsymbol{\div}\sqrt{{a}}}}}=? \\ $$ Answered by Faetmaaa last updated on 27/Feb/24 $$\forall{a}\in\boldsymbol{\mathrm{R}}\backslash\left\{\mathrm{0}\right\} \\ $$$${x}\::=\:\sqrt{\frac{{a}}{\:\sqrt{\frac{{a}}{\:\sqrt{\frac{{a}}{\ldots}}}}}} \\ $$$$\Rightarrow{x}=\sqrt{\frac{{a}}{{x}}} \\…
Question Number 204690 by Davidtim last updated on 25/Feb/24 $$\sqrt[{{p}}]{{a}^{{x}} \boldsymbol{\div}\sqrt[{{q}}]{{b}^{{y}} \boldsymbol{\div}\sqrt[{{r}}]{{c}^{{z}} }}}=? \\ $$ Answered by A5T last updated on 25/Feb/24 $$\left(\frac{{a}^{{x}} }{\left(\frac{{b}^{{y}} }{{c}^{\frac{{z}}{{r}}}…
Question Number 204686 by Davidtim last updated on 25/Feb/24 $$\sqrt{{a}−\sqrt{{a}−\sqrt{{a}−\centerdot\centerdot\centerdot}}}=? \\ $$ Commented by A5T last updated on 25/Feb/24 $${If}\:\sqrt{{a}−\sqrt{{a}−\sqrt{{a}−…}}}\:{converges},\:{then}\:\sqrt{{a}−{x}}={x} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}−{a}=\mathrm{0}\Rightarrow{x}=\frac{−\mathrm{1}\underset{−} {+}\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}} \\…
Question Number 204684 by MrGHK last updated on 25/Feb/24 Commented by mr W last updated on 25/Feb/24 $$\Rightarrow{Q}\mathrm{204522} \\ $$ Commented by Davidtim last updated…
Question Number 204687 by Davidtim last updated on 25/Feb/24 $$\sqrt{{a}−\sqrt{{b}−\sqrt{{c}}}}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 204674 by SEKRET last updated on 25/Feb/24 Commented by A5T last updated on 25/Feb/24 $${O}_{\mathrm{1}} {O}_{\mathrm{2}} \:{is}\:{not}\:{well}-{defined},\:{is}\:{it}\:{a}\:{tangent}\:{line}\:{to} \\ $$$${both}\:{quadrants}? \\ $$ Answered by…