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Category: Differentiation

f-x-x-sin-x-x-g-x-x-Why-is-f-x-0-on-g-x-Are-all-extrema-min-max-inflection-of-f-x-on-g-x-

Question Number 4040 by Filup last updated on 27/Dec/15 $${f}\left({x}\right)={x}^{\mathrm{sin}^{{x}} \left({x}\right)} \\ $$$${g}\left({x}\right)={x} \\ $$$$ \\ $$$$\mathrm{Why}\:\mathrm{is}\:{f}\:'\left({x}\right)=\mathrm{0}\:\mathrm{on}\:{g}\left({x}\right)? \\ $$$$ \\ $$$${Are}\:\boldsymbol{{all}}\:{extrema}\:\left(\mathrm{min},\:\mathrm{max},\:\mathrm{inflection}\right) \\ $$$$\mathrm{of}\:{f}\left({x}\right)\:\mathrm{on}\:{g}\left({x}\right)? \\ $$…

find-the-equation-of-the-circle-which-ends-one-of-the-diameters-of-two-points-p-1-2-3-and-p-2-4-5-

Question Number 69462 by mhmd last updated on 23/Sep/19 $$ \\ $$$${find}\:{the}\:{equation}\:{of}\:{the}\:{circle}\:{which}\:{ends}\:{one}\:{of}\:{the}\:{diameters}\:{of}\:{two}\:{points}\:{p}_{\mathrm{1}} \left(−\mathrm{2},\mathrm{3}\right)\:{and}\:{p}_{\mathrm{2}} \left(\mathrm{4},\mathrm{5}\right) \\ $$ Commented by mathmax by abdo last updated on 23/Sep/19…

find-the-equation-of-the-circle-whose-center-is-the-origin-and-touches-the-line-3x-4y-15-0-

Question Number 69460 by mhmd last updated on 23/Sep/19 $${find}\:{the}\:{equation}\:{of}\:{the}\:{circle}\:{whose}\:{center}\:{is}\:{the}\:{origin}\:{and}\:{touches}\:{the}\:{line}\:\mathrm{3}{x}−\mathrm{4}{y}−\mathrm{15}=\mathrm{0} \\ $$ Commented by mathmax by abdo last updated on 23/Sep/19 $${equation}\:{of}\:{circle}\:{is}\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:={r}^{\mathrm{2}} \:\:{and}\:{r}\:={d}\left(\mathrm{0},{D}\right)…

find-the-value-sin-13pi-6-cos-49pi-4-

Question Number 69458 by mhmd last updated on 23/Sep/19 $${find}\:{the}\:{value}\:{sin}\left(−\mathrm{13}\pi/\mathrm{6}\right)\:,\:{cos}\left(\mathrm{49}\pi/\mathrm{4}\right) \\ $$$$ \\ $$ Commented by kaivan.ahmadi last updated on 23/Sep/19 $${sin}\left(−\mathrm{2}\pi−\frac{\pi}{\mathrm{6}}\right)={sin}\left(−\frac{\pi}{\mathrm{6}}\right)=−{sin}\left(\frac{\pi}{\mathrm{6}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\left(\frac{\mathrm{49}\pi}{\mathrm{4}}\right)={cos}\left(\mathrm{12}\pi+\frac{\pi}{\mathrm{4}}\right)={cos}\left(\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\…

Let-I-z-pi-tan-z-z-4-dz-J-z-pi-cos-z-z-4-dz-and-K-z-pi-cos-Re-z-cos-Im-z-z-4-dz-Show-that-I-J-2-ipi-Show-that-J-K-

Question Number 134998 by snipers237 last updated on 09/Mar/21 $$\:\:\:{Let}\:\:{I}=\:\int_{\mid{z}\mid=\pi} \frac{{tan}\left(\overset{−} {{z}}\right)}{{z}−\mathrm{4}}\:{dz}\:\: \\ $$$$\:{J}=\int_{\mid{z}\mid=\pi} \frac{{cos}\left(\overset{−} {{z}}\right)}{{z}−\mathrm{4}}\:{dz}\:\:\:{and}\:\:{K}=\int_{\mid{z}\mid=\pi} \frac{{cos}\left({Re}\left({z}\right)\right){cos}\left({Im}\left({z}\right)\right)}{{z}−\mathrm{4}}{dz} \\ $$$$\:{Show}\:{that}\:\:{I}={J}\sqrt{\mathrm{2}}=−{i}\pi \\ $$$$\:{Show}\:{that}\:\:{J}={K} \\ $$ Terms of…

find-value-log40-9-4log5-2log6-

Question Number 69457 by mhmd last updated on 23/Sep/19 $${find}\:{value}\:{log}\mathrm{40}/\mathrm{9}\:\:+\mathrm{4}{log}\mathrm{5}\:\:+\mathrm{2}{log}\mathrm{6}\:\:? \\ $$ Answered by MJS last updated on 23/Sep/19 $$\mathrm{log}\:\frac{\mathrm{40}}{\mathrm{9}}\:+\mathrm{4log}\:\mathrm{5}\:+\mathrm{2log}\:\mathrm{6}\:= \\ $$$$=\mathrm{log}\:\mathrm{40}\:−\mathrm{log}\:\mathrm{9}\:+\mathrm{4log}\:\mathrm{5}\:+\mathrm{2log}\:\mathrm{2}\:+\mathrm{2log}\:\mathrm{3}= \\ $$$$=\mathrm{3log}\:\mathrm{2}\:+\mathrm{log}\:\mathrm{5}\:−\mathrm{2log}\:\mathrm{3}\:+\mathrm{4log}\:\mathrm{5}\:+\mathrm{2log}\:\mathrm{2}\:+\mathrm{2log}\:\mathrm{3}= \\…

Question-134878

Question Number 134878 by bemath last updated on 08/Mar/21 Answered by EDWIN88 last updated on 08/Mar/21 $$\mathrm{B}\left(\theta\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{100}.\mathrm{sin}\:\theta\:=\:\mathrm{50}\:\mathrm{sin}\:\theta\: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{radius}\:\mathrm{of}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{r}\:=\:\mathrm{10}.\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right) \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{A}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{100}.\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\right) \\ $$$$\mathrm{A}\left(\theta\right)\:=\:\mathrm{50}\pi\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)…