f-x-x-x-find-d-dx-f-x-and-for-what-x-we-will-get-the-maximum-of-the-function-

Question Number 11546 by Nayon last updated on 28/Mar/17
$$\\$$$$\\$$$$\\$$$$\\$$$$\:\:\:\:\:\:\:\:{f}\left({x}\right)=^{{x}} \sqrt{{x}}\:{find}\:\frac{{d}}{{dx}}\left({f}\left({x}\right)\right)\: \\$$$$\:\:\:\:\:\:\:\:\:\:\:{and}\:{for}\:{what}\:{x}\:,{we}\:{will}\:{get}\:{the}\: \\$$$$\:{maximum}\:{of}\:{the}\:{function}..? \\$$$$\\$$$$\\$$$$\:\:\:\: \\$$
Answered by sma3l2996 last updated on 28/Mar/17
$$\frac{{d}\left({f}\left({x}\right)\right)}{{dx}}=\frac{{d}\left(\sqrt[{{x}}]{{x}}\right)}{{dx}}=\frac{{d}\left({e}^{{ln}\left({x}^{\frac{\mathrm{1}}{{x}}} \right)} \right)}{{dx}}=\frac{{d}\left({e}^{\frac{{ln}\left({x}\right)}{{x}}} \right)}{{dx}} \\$$$$=\frac{{d}\left(\frac{{ln}\left({x}\right)}{{x}}\right)}{{dx}}{e}^{\frac{{ln}\left({x}\right)}{{x}}} =\frac{\mathrm{1}−{ln}\left({x}\right)}{{x}^{\mathrm{2}} }{e}^{{ln}\left(\sqrt[{{x}}]{{x}}\right)} \\$$$$=\frac{\left(\mathrm{1}−{ln}\left({x}\right)\right)\sqrt[{{x}}]{{x}}}{{x}^{\mathrm{2}} } \\$$$${when}\:{f}\:{get}\:{the}\:{maximum}\:{so}\:{f}'\left({x}\right)=\mathrm{0} \\$$$$\frac{\left(\mathrm{1}−{ln}\left({x}\right)\right)\sqrt[{{x}}]{{x}}}{{x}^{\mathrm{2}} }=\mathrm{0}\:\:{and}\:\:\sqrt[{{x}}]{{x}}={e}^{\frac{{ln}\left({x}\right)}{{x}}} >\mathrm{0} \\$$$$\mathrm{1}−{ln}\left({x}\right)=\mathrm{0}\:\Leftrightarrow{ln}\left({x}\right)=\mathrm{1}\: \\$$$${x}={e}\: \\$$
Commented by Nayon last updated on 28/Mar/17
$${thanks}\:{a}\:{lot}\: \\$$
Answered by Joel576 last updated on 28/Mar/17
$${y}\:=\:{x}^{\frac{\mathrm{1}}{{x}}} \\$$$$\mathrm{ln}\:{y}\:=\:\frac{\mathrm{1}}{{x}}\:.\:\mathrm{ln}\:{x} \\$$$$\frac{{d}}{{dy}}\:\left(\mathrm{ln}\:{y}\right)\:=\:\frac{{d}}{{dx}}\:\left(\frac{\mathrm{1}}{{x}}\:.\:\mathrm{ln}\:{x}\right) \\$$$$\frac{{dy}}{{dx}}\:\frac{\mathrm{1}}{{y}}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}\:−\:\frac{\mathrm{ln}\:{x}}{{x}^{\mathrm{2}} } \\$$$$\frac{{dy}}{{dx}}\:\frac{\mathrm{1}}{{y}}\:=\:\frac{\mathrm{1}\:−\:\mathrm{ln}\:{x}}{{x}^{\mathrm{2}} } \\$$$$\frac{{dy}}{{dx}}\:=\:{y}\left(\frac{\mathrm{1}\:−\:\mathrm{ln}\:{x}}{{x}^{\mathrm{2}} }\right) \\$$$${f}\:'\left({x}\right)\:=\:\frac{{dy}}{{dx}}\:=\:\frac{{x}^{\frac{\mathrm{1}}{{x}}} \:\left(\mathrm{1}\:−\:\mathrm{ln}\:{x}\right)}{{x}^{\mathrm{2}} } \\$$$$\\$$$$\mathrm{Function}\:{f}\left({x}\right)\:\mathrm{will}\:\mathrm{be}\:\mathrm{maximum}\:\mathrm{if}\:{f}\:'\left({x}\right)\:=\:\mathrm{0} \\$$$$\frac{{x}^{\frac{\mathrm{1}}{{x}}} \:\left(\mathrm{1}\:−\:\mathrm{ln}\:{x}\right)}{{x}^{\mathrm{2}} }\:=\:\mathrm{0} \\$$$$\bullet\:{x}^{\frac{\mathrm{1}}{{x}}} \:=\:\mathrm{0}\:\:\:\rightarrow\:\mathrm{undefined}\:\left(?\right) \\$$$$\bullet\:\mathrm{1}\:−\:\mathrm{ln}\:{x}\:=\:\mathrm{0} \\$$$$\:\:\:\:\mathrm{ln}\:{x}\:=\:\mathrm{1} \\$$$$\:\:\:\:{x}\:=\:{e} \\$$
Commented by Nayon last updated on 28/Mar/17
$${in}\:{the}\:{third}\:{line}\:\frac{{d}}{{dx}}\left({lny}\right)=\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{x}}{lnx}\right) \\$$$${or}\:\frac{{d}}{{dy}}\left({lny}\right)=\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{x}}.{lnx}\right)???????? \\$$
Commented by Joel576 last updated on 28/Mar/17
$$\mathrm{Differential}\:\mathrm{at}\:\mathrm{both}\:\mathrm{section} \\$$$$\frac{{d}}{{dy}}\:\left(\mathrm{ln}\:{y}\right)\:=\:\frac{{d}}{{dx}}\:\left(\frac{\mathrm{1}}{{x}}\:.\:\mathrm{ln}\:{x}\right) \\$$
Commented by Nayon last updated on 28/Mar/17
$${wrong}\:{it}'{ll}\:{be}\:\:\frac{{d}}{{dy}}{lny}.\frac{{dy}}{{dx}} \\$$