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Category: Differentiation

y-x-3-x-5-2-x-2-x-2-Find-local-maxima-and-minima-hence-draw-the-graph-Also-find-radius-of-circle-touching-all-three-sections-of-the-curve-that-results-

Question Number 36341 by ajfour last updated on 02/Jun/18 $${y}=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{5}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}+\mathrm{2}\right)} \\ $$$${Find}\:{local}\:{maxima}\:{and}\:{minima}; \\ $$$${hence}\:{draw}\:{the}\:{graph}.{Also}\:{find} \\ $$$${radius}\:{of}\:{circle}\:{touching}\:{all}\:{three} \\ $$$${sections}\:{of}\:{the}\:{curve}\:{that}\:{results}. \\ $$$$ \\ $$ Answered…

Question-167404

Question Number 167404 by bharath23 last updated on 15/Mar/22 Answered by alephzero last updated on 15/Mar/22 $$\frac{\partial{f}}{\partial{x}}\:=\:\frac{\partial}{\partial{x}}\left({y}\:\mathrm{cos}\:{xy}\right)\:=\:{y}\:\frac{\partial}{\partial{x}}\left(\mathrm{cos}\:{xy}\right) \\ $$$$=\:{y}\:\frac{\partial\left(\mathrm{cos}\:{xy}\right)}{\partial\left({xy}\right)}\:\centerdot\:\frac{\partial\left({xy}\right)}{\partial{x}}\:=\:−{y}^{\mathrm{2}} \:\mathrm{sin}\:{xy}\:\blacksquare \\ $$ Terms of Service…

Find-domain-and-range-of-the-function-f-x-x-2-6x-8-x-5-Also-draw-the-graph-

Question Number 36316 by ajfour last updated on 31/May/18 $${Find}\:{domain}\:{and}\:{range}\:{of}\:{the} \\ $$$${function}\:\:{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{8}}{{x}−\mathrm{5}}\:\:. \\ $$$${Also}\:{draw}\:{the}\:{graph}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 31/May/18 $${y}=\frac{{x}^{\mathrm{2}}…

Question-167301

Question Number 167301 by mnjuly1970 last updated on 12/Mar/22 Answered by amin96 last updated on 12/Mar/22 $$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}−\boldsymbol{\mathrm{cos}}\left(\frac{\pi}{\mathrm{4}}−\boldsymbol{\mathrm{x}}\right)\right)}\boldsymbol{\mathrm{dx}}\:\:\:\:\frac{\pi}{\mathrm{4}}−\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{t}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{t}}\left[\frac{\pi}{\mathrm{8}};\:\frac{\pi}{\mathrm{4}}\right]\:\:\:\:\boldsymbol{{I}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\frac{\pi}{\mathrm{8}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\boldsymbol{\mathrm{dt}}}{\left(\mathrm{1}−\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{t}}\right)\right)}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\frac{\pi}{\mathrm{8}}}…