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Question Number 11546 by Nayon last updated on 28/Mar/17

f(x)=^x (√x) find (d/dx)(f(x)) and for what x ,we will get the maximum of the function..?

$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:{f}\left({x}\right)=^{{x}} \sqrt{{x}}\:{find}\:\frac{{d}}{{dx}}\left({f}\left({x}\right)\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{and}\:{for}\:{what}\:{x}\:,{we}\:{will}\:{get}\:{the}\: \\ $$$$\:{maximum}\:{of}\:{the}\:{function}..? \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\: \\ $$

Answered by sma3l2996 last updated on 28/Mar/17

((d(f(x)))/dx)=((d((x)^(1/x) ))/dx)=((d(e^(ln(x^(1/x) )) ))/dx)=((d(e^((ln(x))/x) ))/dx) =((d(((ln(x))/x)))/dx)e^((ln(x))/x) =((1−ln(x))/x^2 )e^(ln((x)^(1/x) )) =(((1−ln(x))(x)^(1/x) )/x^2 ) when f get the maximum so f′(x)=0 (((1−ln(x))(x)^(1/x) )/x^2 )=0 and (x)^(1/x) =e^((ln(x))/x) >0 1−ln(x)=0 ⇔ln(x)=1 x=e

$$\frac{{d}\left({f}\left({x}\right)\right)}{{dx}}=\frac{{d}\left(\sqrt[{{x}}]{{x}}\right)}{{dx}}=\frac{{d}\left({e}^{{ln}\left({x}^{\frac{\mathrm{1}}{{x}}} \right)} \right)}{{dx}}=\frac{{d}\left({e}^{\frac{{ln}\left({x}\right)}{{x}}} \right)}{{dx}} \\ $$$$=\frac{{d}\left(\frac{{ln}\left({x}\right)}{{x}}\right)}{{dx}}{e}^{\frac{{ln}\left({x}\right)}{{x}}} =\frac{\mathrm{1}−{ln}\left({x}\right)}{{x}^{\mathrm{2}} }{e}^{{ln}\left(\sqrt[{{x}}]{{x}}\right)} \\ $$$$=\frac{\left(\mathrm{1}−{ln}\left({x}\right)\right)\sqrt[{{x}}]{{x}}}{{x}^{\mathrm{2}} } \\ $$$${when}\:{f}\:{get}\:{the}\:{maximum}\:{so}\:{f}'\left({x}\right)=\mathrm{0} \\ $$$$\frac{\left(\mathrm{1}−{ln}\left({x}\right)\right)\sqrt[{{x}}]{{x}}}{{x}^{\mathrm{2}} }=\mathrm{0}\:\:{and}\:\:\sqrt[{{x}}]{{x}}={e}^{\frac{{ln}\left({x}\right)}{{x}}} >\mathrm{0} \\ $$$$\mathrm{1}−{ln}\left({x}\right)=\mathrm{0}\:\Leftrightarrow{ln}\left({x}\right)=\mathrm{1}\: \\ $$$${x}={e}\: \\ $$

Commented by Nayon last updated on 28/Mar/17

thanks a lot

$${thanks}\:{a}\:{lot}\: \\ $$

Answered by Joel576 last updated on 28/Mar/17

y = x^(1/x) ln y = (1/x) . ln x (d/dy) (ln y) = (d/dx) ((1/x) . ln x) (dy/dx) (1/y) = (1/(x^2 )) − ((ln x)/x^2 ) (dy/dx) (1/y) = ((1 − ln x)/x^2 ) (dy/dx) = y(((1 − ln x)/x^2 )) f ′(x) = (dy/dx) = ((x^(1/x) (1 − ln x))/x^2 ) Function f(x) will be maximum if f ′(x) = 0 ((x^(1/x) (1 − ln x))/x^2 ) = 0 • x^(1/x) = 0 → undefined (?) • 1 − ln x = 0 ln x = 1 x = e

$${y}\:=\:{x}^{\frac{\mathrm{1}}{{x}}} \\ $$$$\mathrm{ln}\:{y}\:=\:\frac{\mathrm{1}}{{x}}\:.\:\mathrm{ln}\:{x} \\ $$$$\frac{{d}}{{dy}}\:\left(\mathrm{ln}\:{y}\right)\:=\:\frac{{d}}{{dx}}\:\left(\frac{\mathrm{1}}{{x}}\:.\:\mathrm{ln}\:{x}\right) \\ $$$$\frac{{dy}}{{dx}}\:\frac{\mathrm{1}}{{y}}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}\:−\:\frac{\mathrm{ln}\:{x}}{{x}^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}\:\frac{\mathrm{1}}{{y}}\:=\:\frac{\mathrm{1}\:−\:\mathrm{ln}\:{x}}{{x}^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}\:=\:{y}\left(\frac{\mathrm{1}\:−\:\mathrm{ln}\:{x}}{{x}^{\mathrm{2}} }\right) \\ $$$${f}\:'\left({x}\right)\:=\:\frac{{dy}}{{dx}}\:=\:\frac{{x}^{\frac{\mathrm{1}}{{x}}} \:\left(\mathrm{1}\:−\:\mathrm{ln}\:{x}\right)}{{x}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{Function}\:{f}\left({x}\right)\:\mathrm{will}\:\mathrm{be}\:\mathrm{maximum}\:\mathrm{if}\:{f}\:'\left({x}\right)\:=\:\mathrm{0} \\ $$$$\frac{{x}^{\frac{\mathrm{1}}{{x}}} \:\left(\mathrm{1}\:−\:\mathrm{ln}\:{x}\right)}{{x}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\bullet\:{x}^{\frac{\mathrm{1}}{{x}}} \:=\:\mathrm{0}\:\:\:\rightarrow\:\mathrm{undefined}\:\left(?\right) \\ $$$$\bullet\:\mathrm{1}\:−\:\mathrm{ln}\:{x}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\mathrm{ln}\:{x}\:=\:\mathrm{1} \\ $$$$\:\:\:\:{x}\:=\:{e} \\ $$

Commented by Nayon last updated on 28/Mar/17

in the third line (d/dx)(lny)=(d/dx)((1/x)lnx) or (d/dy)(lny)=(d/dx)((1/x).lnx)????????

$${in}\:{the}\:{third}\:{line}\:\frac{{d}}{{dx}}\left({lny}\right)=\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{x}}{lnx}\right) \\ $$$${or}\:\frac{{d}}{{dy}}\left({lny}\right)=\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{x}}.{lnx}\right)???????? \\ $$

Commented by Joel576 last updated on 28/Mar/17

Differential at both section (d/dy) (ln y) = (d/dx) ((1/x) . ln x)

$$\mathrm{Differential}\:\mathrm{at}\:\mathrm{both}\:\mathrm{section} \\ $$$$\frac{{d}}{{dy}}\:\left(\mathrm{ln}\:{y}\right)\:=\:\frac{{d}}{{dx}}\:\left(\frac{\mathrm{1}}{{x}}\:.\:\mathrm{ln}\:{x}\right) \\ $$

Commented by Nayon last updated on 28/Mar/17

wrong it′ll be (d/dy)lny.(dy/dx)

$${wrong}\:{it}'{ll}\:{be}\:\:\frac{{d}}{{dy}}{lny}.\frac{{dy}}{{dx}} \\ $$

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