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Category: Differentiation

Question-65367

Question Number 65367 by ajfour last updated on 29/Jul/19 Commented by ajfour last updated on 29/Jul/19 $$\mathrm{For}\:\mathrm{maximum}\:\mathrm{arc}\:\mathrm{length}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{origin}\:\mathrm{centered}\:\mathrm{circle}\:\mathrm{within} \\ $$$$\mathrm{the}\:\mathrm{shown}\:\mathrm{unit}\:\mathrm{circle},\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$…

let-f-x-y-x-2-xy-x-2-y-2-prove-from-defination-of-derivative-that-Df-x-y-1-2-4x-y-2x-4y-

Question Number 130884 by LYKA last updated on 30/Jan/21 $${let}\:{f}\left({x}.{y}\right)=\left[{x}^{\mathrm{2}} +{xy},{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right]\: \\ $$$$ \\ $$$${prove}\:{from}\:{defination}\:{of}\: \\ $$$${derivative}\:{that}\:: \\ $$$${Df}_{\left({x},{y}\right)} \left(\mathrm{1},\mathrm{2}\right)=\left[\mathrm{4}\boldsymbol{{x}}+\boldsymbol{{y}},−\mathrm{2}{x}+\mathrm{4}{y}\right] \\ $$$$ \\…

Question-65300

Question Number 65300 by rajesh4661kumar@gmail.com last updated on 28/Jul/19 Commented by kaivan.ahmadi last updated on 28/Jul/19 $${f}\left({x}+\Delta{x}\right)={f}\left({x}\right)+{f}'\left({x}\right).\Delta{x} \\ $$$${f}\left({x}\right)={logx}\Rightarrow{f}'\left({x}\right)=\frac{\mathrm{1}}{{xln}\mathrm{10}} \\ $$$$\left.{f}\left(\mathrm{10}.\mathrm{02}\right)\right)={f}\left(\mathrm{10}+\mathrm{0}.\mathrm{02}\right)={f}\left(\mathrm{10}\right)+{f}'\left(\mathrm{10}\right)×\mathrm{0}.\mathrm{02}= \\ $$$$\mathrm{2}.\mathrm{3026}+\frac{\mathrm{1}}{\mathrm{10}{ln}\mathrm{10}}×\mathrm{0}.\mathrm{02}=\mathrm{2}.\mathrm{3026}+\frac{\mathrm{2}}{\mathrm{1000}{ln}\mathrm{10}} \\ $$$$…

Question-130780

Question Number 130780 by LYKA last updated on 28/Jan/21 Commented by kaivan.ahmadi last updated on 29/Jan/21 $$\overset{\rightarrow} {{b}}=\left(\mathrm{2},−\mathrm{1},\mathrm{1}\right)\Rightarrow\mid\overset{\rightarrow} {{b}}\mid=\sqrt{\mathrm{6}}\Rightarrow\overset{\rightarrow} {{u}}=\frac{\overset{\rightarrow} {{b}}}{\mid\overset{\rightarrow} {{b}}\mid}=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}},\frac{−\mathrm{1}}{\:\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\right) \\ $$$$\bigtriangledown{f}=\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{yz},\mathrm{3}{y}^{\mathrm{2}}…