Differentiating $f(x)= sqrt[x]{x}$ from first principles












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$begingroup$


An interesting fact is that the point of maximum of $y=sqrt[x]{x}$ is attained when x equals to exactly e (as seen from the pic)



I know how to differentiate $y= sqrt[x]{x}$, by logarithmic methodology, but can anyone teach me how to differentiate this function using only the first principle?



Note: I am still a beginner, thus it would help greatly if you can include all the steps.



enter image description here






calculus







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  • $begingroup$
    In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
    $endgroup$
    – ComFreek
    Dec 31 '18 at 15:34






  • 1




    $begingroup$
    What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
    $endgroup$
    – Bernard Massé
    Dec 31 '18 at 16:12










  • $begingroup$
    What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
    $endgroup$
    – egreg
    Jan 1 at 10:46


















3












$begingroup$


An interesting fact is that the point of maximum of $y=sqrt[x]{x}$ is attained when x equals to exactly e (as seen from the pic)



I know how to differentiate $y= sqrt[x]{x}$, by logarithmic methodology, but can anyone teach me how to differentiate this function using only the first principle?



Note: I am still a beginner, thus it would help greatly if you can include all the steps.



enter image description here






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
    $endgroup$
    – ComFreek
    Dec 31 '18 at 15:34






  • 1




    $begingroup$
    What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
    $endgroup$
    – Bernard Massé
    Dec 31 '18 at 16:12










  • $begingroup$
    What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
    $endgroup$
    – egreg
    Jan 1 at 10:46
















3












3








3


1



$begingroup$


An interesting fact is that the point of maximum of $y=sqrt[x]{x}$ is attained when x equals to exactly e (as seen from the pic)



I know how to differentiate $y= sqrt[x]{x}$, by logarithmic methodology, but can anyone teach me how to differentiate this function using only the first principle?



Note: I am still a beginner, thus it would help greatly if you can include all the steps.



enter image description here






calculus







share|cite|improve this question











$endgroup$




An interesting fact is that the point of maximum of $y=sqrt[x]{x}$ is attained when x equals to exactly e (as seen from the pic)



I know how to differentiate $y= sqrt[x]{x}$, by logarithmic methodology, but can anyone teach me how to differentiate this function using only the first principle?



Note: I am still a beginner, thus it would help greatly if you can include all the steps.



enter image description here






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 15:59









Namaste

1




1










asked Dec 31 '18 at 15:26







user630471



















  • $begingroup$
    In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
    $endgroup$
    – ComFreek
    Dec 31 '18 at 15:34






  • 1




    $begingroup$
    What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
    $endgroup$
    – Bernard Massé
    Dec 31 '18 at 16:12










  • $begingroup$
    What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
    $endgroup$
    – egreg
    Jan 1 at 10:46




















  • $begingroup$
    In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
    $endgroup$
    – ComFreek
    Dec 31 '18 at 15:34






  • 1




    $begingroup$
    What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
    $endgroup$
    – Bernard Massé
    Dec 31 '18 at 16:12










  • $begingroup$
    What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
    $endgroup$
    – egreg
    Jan 1 at 10:46


















$begingroup$
In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
$endgroup$
– ComFreek
Dec 31 '18 at 15:34




$begingroup$
In addition to the answers given so far you should also verify that it is indeed a local maximum point, e.g. by showing $f''(e) < 0$ or that $f'$ changes signs at $e$.
$endgroup$
– ComFreek
Dec 31 '18 at 15:34




1




1




$begingroup$
What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
$endgroup$
– Bernard Massé
Dec 31 '18 at 16:12




$begingroup$
What do you mean by "using only the first principle"? If you mean by using the definition, I think you're in for a rather rough ride.
$endgroup$
– Bernard Massé
Dec 31 '18 at 16:12












$begingroup$
What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
$endgroup$
– egreg
Jan 1 at 10:46






$begingroup$
What's the purpose of going through a very complicated manipulation that's essentially nothing else than “reproving” the chain rule and the product rule in a special case? I'll reveal you a secret: knowing how to differentiate gives you knowledge of a very wide range of limits! Many basic exercises on limits are just “adorned” derivatives.
$endgroup$
– egreg
Jan 1 at 10:46












3 Answers
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$begingroup$

A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$



Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$






share|cite|improve this answer









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    6












    $begingroup$

    I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
      $$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
      The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
      $$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

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        active

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        votes









        10












        $begingroup$

        A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
        This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$



        Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$






        share|cite|improve this answer









        $endgroup$


















          10












          $begingroup$

          A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
          This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$



          Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$






          share|cite|improve this answer









          $endgroup$
















            10












            10








            10





            $begingroup$

            A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
            This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$



            Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$






            share|cite|improve this answer









            $endgroup$



            A nice trick is to rewrite $y=sqrt[x]x$ as $$y=e^{frac{ln(x)}{x}}tag{1}$$
            This is because $$y=sqrt[x]x=x^frac{1}{x}=e^{lnleft(x^frac{1}{x}right)}=e^frac{ln(x)}{x}$$



            Differentiating $(1)$, and setting it equal to $0$ yields: $$frac{dy}{dx}=e^{frac{ln(x)}{x}}left(frac{1}{x^2}-frac{ln(x)}{x^2}right)=0\Rightarrow frac{1}{x^2}-frac{ln(x)}{x^2}=0\Rightarrow ln(x)=1\Rightarrow x=e$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 31 '18 at 15:32









            cansomeonehelpmeoutcansomeonehelpmeout

            7,3193935




            7,3193935























                6












                $begingroup$

                I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.






                share|cite|improve this answer









                $endgroup$


















                  6












                  $begingroup$

                  I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.






                  share|cite|improve this answer









                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.






                    share|cite|improve this answer









                    $endgroup$



                    I'd consider the case $x>0$. Since the functions $exp:{Bbb R}rightarrow {Bbb R}_{>0}: xmapsto e^x$ and $ln:{Bbb R}_{>0}rightarrow {Bbb R}: xmapsto ln(x)$ are inverse to each other, we have $sqrt[x]x = x^{1/x} = e^{ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 31 '18 at 15:31









                    WuestenfuxWuestenfux

                    5,3331513




                    5,3331513























                        1












                        $begingroup$

                        You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
                        $$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
                        The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
                        $$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
                          $$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
                          The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
                          $$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
                            $$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
                            The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
                            $$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$






                            share|cite|improve this answer









                            $endgroup$



                            You have the "outer" function $$Psi(u,v):=u^vqquad(uin{mathbb R}_{>0}, vin{mathbb R})$$ of two variables, with partial derivatives
                            $$Psi_u(u,v)=v,u^{v-1}>,qquadPsi_v(u,v)=log (u) u^v .$$
                            The function $$f(x):=Psileft(x,{1over x}right)$$ then has to be differentiated using the chain rule:
                            $$eqalign{f'(x)&=Psi_u(x,1/x)cdot 1+Psi_v(x,1/x)cdotleft(-{1over x^2}right)cr &={1over x}x^{1/x-1}+log (x)>x^{1/x}left(-{1over x^2}right)cr &={root x of xover x^2}bigl(1-log (x)bigr) .cr}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 1 at 10:35









                            Christian BlatterChristian Blatter

                            175k8115327




                            175k8115327






























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