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Category: Differentiation

Question-163367

Question Number 163367 by poeter last updated on 06/Jan/22 Answered by som(math1967) last updated on 06/Jan/22 $$\mathrm{1}.\:{equn}\:{of}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{10}\:\left({given}\right) \\ $$$$\:\:\frac{{dy}}{{dx}}=\frac{−{x}}{{y}} \\ $$$$\left[\frac{{dy}}{{dx}}\right]_{\mathrm{1},\mathrm{3}} =\frac{−\mathrm{1}}{\mathrm{3}} \\…

put-gcd-a-b-a-b-if-a-b-a-c-b-c-1-prove-that-abc-ab-ac-bc-1-

Question Number 163271 by mnjuly1970 last updated on 05/Jan/22 $$ \\ $$$$\:\:\:\:\:\:{put}\::\:\:{gcd}\left(\:{a}\:,\:{b}\:\right)=\:\left({a},\:{b}\:\right) \\ $$$$\:\:\:\:\:\:\:{if}\:\:\:\left(\:{a}\:,{b}\:\right)=\:\left({a}\:,{c}\:\right)=\:\left({b}\:,{c}\:\right)=\mathrm{1} \\ $$$${prove}\:{that}\::\:\:\left({abc}\:,\:{ab}\:+{ac}\:+{bc}\:\right)=\mathrm{1} \\ $$$$ \\ $$ Terms of Service Privacy Policy…

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Question Number 163134 by mnjuly1970 last updated on 04/Jan/22 $$ \\ $$$$\:\:\:\:{prove}\:\:{or}\:{disprove} \\ $$$$ \\ $$$$\:\:\:\:\int_{\mathrm{2}\pi} ^{\:\mathrm{4}\pi} \frac{\:{sin}\left({x}\right)}{{x}}\:{dx}\:>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{because} \\ $$$$\:\int_{\mathrm{2}\pi} ^{\:\mathrm{3}\pi} \frac{\:{sin}\left({x}\:\right)}{{x}}\:{dx}\:>\:\int_{\mathrm{3}\pi} ^{\:\mathrm{4}\pi}…

F-x-x-2-4x-2-1-3-local-maximum-absolut-maximum-

Question Number 163080 by tounghoungko last updated on 03/Jan/22 $$\:\:\:\:\:\:\:{F}\left({x}\right)=\:\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\left.\begin{matrix}{{local}\:{maximum}}\\{{absolut}\:{maximum}}\end{matrix}\right\}\:=? \\ $$ Answered by mahdipoor last updated on 03/Jan/22 $${F}\:'=\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\frac{−\mathrm{1}}{\mathrm{3}}}…

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Question Number 163045 by mnjuly1970 last updated on 03/Jan/22 $$ \\ $$$$\:\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(\:\mathrm{2}{n}+\mathrm{1}\:\right)!!}{\left(\mathrm{2}{n}\:\right)!!}\:\frac{\mathrm{1}}{\mathrm{2}^{\:{n}} \left(\mathrm{2}{n}\:+\mathrm{1}\right)^{\:\mathrm{2}} }\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}−\mathrm{1} \\ $$ Answered by qaz…