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Category: Differentiation

Question-188387

Question Number 188387 by cortano12 last updated on 28/Feb/23 Answered by Frix last updated on 28/Feb/23 $$\mathrm{I}\:\mathrm{think}\:\mathrm{because}\:\mathrm{of}\:\mathrm{symmetry}\:{a}={b}={c}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{Minimum}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Answered by mnjuly1970 last…

general-calculus-i-1-1-4x-n-0-2n-n-x-n-ii-pi-2-n-0-2n-n-4-n-1-2n-1-

Question Number 122704 by mnjuly1970 last updated on 19/Nov/20 $$\:\:\:\:\:\:\:\:\:\:\:…\:{general}\:\:{calculus}… \\ $$$$\:{i}:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}}}\:\overset{??} {=}\underset{{n}=\mathrm{0}\:} {\overset{\infty} {\sum}}\left[\begin{pmatrix}{\mathrm{2}{n}}\\{\:{n}}\end{pmatrix}\:\:{x}^{{n}} \right] \\ $$$$\:{ii}:\:\frac{\pi}{\mathrm{2}}\:\overset{?} {=}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{\:{n}}\end{pmatrix}}{\mathrm{4}^{{n}} }\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\: \\…

Question-57084

Question Number 57084 by frimpshaddie last updated on 30/Mar/19 Commented by Kunal12588 last updated on 30/Mar/19 $$\left({d}\right)\:{when}\:{the}\:{velocity}\:{of}\:{particle}\:{becomes}\:\mathrm{0} \\ $$$${then}\:{the}\:{particle}\:{is}\:{said}\:{to}\:{be}\:{in}\:{rest} \\ $$$$\left({horizontal}\:{motion}\:{only}\right) \\ $$$$\therefore{v}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{t}^{\mathrm{2}}…

nice-calculus-prove-that-0-1-ln-1-x-2-x-dx-pi-2-12-m-n-1970-

Question Number 122525 by mnjuly1970 last updated on 17/Nov/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:{prove}\:{that}\:\:\::::::\gg\:\:\:\Psi=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}−{x}}{dx}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:…\:{m}.{n}.\mathrm{1970}… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by Dwaipayan Shikari…

find-dy-dx-y-2x-x-

Question Number 188037 by Michaelfaraday last updated on 25/Feb/23 $${find}\:\frac{{dy}}{{dx}} \\ $$$${y}=\mathrm{2}{x}^{\sqrt{{x}}} \\ $$ Answered by horsebrand11 last updated on 25/Feb/23 $$\:\mathrm{ln}\:{y}=\sqrt{{x}}\:\mathrm{ln}\:\left(\mathrm{2}{x}\right) \\ $$$$\:\frac{\mathrm{1}}{{y}}.{y}'=\frac{\mathrm{ln}\:\left(\mathrm{2}{x}\right)}{\mathrm{2}\sqrt{{x}}}+\frac{\mathrm{2}\sqrt{{x}}}{\mathrm{2}{x}} \\…