Use logarithmic differentiation to find $frac{dy}{dx}$ of $y=(ln x)^{ln x}$












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$begingroup$


I've been asked to find the derivative of $ln x^{ln x}$ using a method called "logarithmic differentiation". I'm not familiar with this method of differentiation, and a search on Wikipedia doesn't tell me anything I can understand. I know that $frac d{dx}ln x=frac1x$, the Inverse Function Theorem $left(frac d{dx}f^{-1}{'}=frac1{f'(f^{-1}(x))}right)$ and the Chain, Addition, Subtraction, Product and Quotient Rules. Can anybody help me with this question?






calculus derivatives logarithms







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  • 2




    $begingroup$
    Hint: Take the natural logarithm on both sides of the equation which gives you $ln (y)=ln(x) ln(ln (x))$. Now all you have to do is to implicitly differentiate and use the product rule and the chain rule.
    $endgroup$
    – Paras Khosla
    Feb 1 at 8:48










  • $begingroup$
    The use of logarithmic differentiation replaces the product rule, the quotient rule, the power rule. Think about it and remember.
    $endgroup$
    – Claude Leibovici
    Feb 1 at 9:52
















2












$begingroup$


I've been asked to find the derivative of $ln x^{ln x}$ using a method called "logarithmic differentiation". I'm not familiar with this method of differentiation, and a search on Wikipedia doesn't tell me anything I can understand. I know that $frac d{dx}ln x=frac1x$, the Inverse Function Theorem $left(frac d{dx}f^{-1}{'}=frac1{f'(f^{-1}(x))}right)$ and the Chain, Addition, Subtraction, Product and Quotient Rules. Can anybody help me with this question?






calculus derivatives logarithms







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: Take the natural logarithm on both sides of the equation which gives you $ln (y)=ln(x) ln(ln (x))$. Now all you have to do is to implicitly differentiate and use the product rule and the chain rule.
    $endgroup$
    – Paras Khosla
    Feb 1 at 8:48










  • $begingroup$
    The use of logarithmic differentiation replaces the product rule, the quotient rule, the power rule. Think about it and remember.
    $endgroup$
    – Claude Leibovici
    Feb 1 at 9:52














2












2








2





$begingroup$


I've been asked to find the derivative of $ln x^{ln x}$ using a method called "logarithmic differentiation". I'm not familiar with this method of differentiation, and a search on Wikipedia doesn't tell me anything I can understand. I know that $frac d{dx}ln x=frac1x$, the Inverse Function Theorem $left(frac d{dx}f^{-1}{'}=frac1{f'(f^{-1}(x))}right)$ and the Chain, Addition, Subtraction, Product and Quotient Rules. Can anybody help me with this question?






calculus derivatives logarithms







share|cite|improve this question











$endgroup$




I've been asked to find the derivative of $ln x^{ln x}$ using a method called "logarithmic differentiation". I'm not familiar with this method of differentiation, and a search on Wikipedia doesn't tell me anything I can understand. I know that $frac d{dx}ln x=frac1x$, the Inverse Function Theorem $left(frac d{dx}f^{-1}{'}=frac1{f'(f^{-1}(x))}right)$ and the Chain, Addition, Subtraction, Product and Quotient Rules. Can anybody help me with this question?






calculus derivatives logarithms




calculus derivatives logarithms






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edited Feb 1 at 9:35









Bernard

124k741117




124k741117










asked Feb 1 at 8:41









KykyKyky

470414




470414








  • 2




    $begingroup$
    Hint: Take the natural logarithm on both sides of the equation which gives you $ln (y)=ln(x) ln(ln (x))$. Now all you have to do is to implicitly differentiate and use the product rule and the chain rule.
    $endgroup$
    – Paras Khosla
    Feb 1 at 8:48










  • $begingroup$
    The use of logarithmic differentiation replaces the product rule, the quotient rule, the power rule. Think about it and remember.
    $endgroup$
    – Claude Leibovici
    Feb 1 at 9:52














  • 2




    $begingroup$
    Hint: Take the natural logarithm on both sides of the equation which gives you $ln (y)=ln(x) ln(ln (x))$. Now all you have to do is to implicitly differentiate and use the product rule and the chain rule.
    $endgroup$
    – Paras Khosla
    Feb 1 at 8:48










  • $begingroup$
    The use of logarithmic differentiation replaces the product rule, the quotient rule, the power rule. Think about it and remember.
    $endgroup$
    – Claude Leibovici
    Feb 1 at 9:52








2




2




$begingroup$
Hint: Take the natural logarithm on both sides of the equation which gives you $ln (y)=ln(x) ln(ln (x))$. Now all you have to do is to implicitly differentiate and use the product rule and the chain rule.
$endgroup$
– Paras Khosla
Feb 1 at 8:48




$begingroup$
Hint: Take the natural logarithm on both sides of the equation which gives you $ln (y)=ln(x) ln(ln (x))$. Now all you have to do is to implicitly differentiate and use the product rule and the chain rule.
$endgroup$
– Paras Khosla
Feb 1 at 8:48












$begingroup$
The use of logarithmic differentiation replaces the product rule, the quotient rule, the power rule. Think about it and remember.
$endgroup$
– Claude Leibovici
Feb 1 at 9:52




$begingroup$
The use of logarithmic differentiation replaces the product rule, the quotient rule, the power rule. Think about it and remember.
$endgroup$
– Claude Leibovici
Feb 1 at 9:52










3 Answers
3






active

oldest

votes


















3












$begingroup$

If $y = ln(x)^{ln(x)}$, then $ln(y) = lnleft(ln(x)^{ln(x)}right) = ln(x) cdot ln(ln(x))$. The right hand side is straightforward to differentiate, using chain and product rules. I'll leave that to you.



The left side can be differentiated with the chain rule, remembering that $y$ is a function of $x$. We do the usual: differentiate the outer function $ln$, leaving the inner function $y$ in tact, then multiply by the derivative of the inner function. That is,
$$frac{mathrm{d}}{mathrm{d}x} ln(y) = frac{1}{y} cdot y' = frac{y'}{ln(x)^{ln(x)}}$$
So, differentiating both sides yields
$$frac{y'}{ln(x)^{ln(x)}} = frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)),$$
hence
$$y' = ln(x)^{ln(x)} cdot frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I now understand how logarithmic differentiation works after looking at your example, thanks!
    $endgroup$
    – Kyky
    Feb 1 at 9:33



















2












$begingroup$

$$y=(ln(x))^{ln(x)}$$
$$Longrightarrow ln(y)=ln(x)ln(ln(x))$$
$$Longrightarrow frac{y'}{y}=frac{ln(x)}{xln(x)}+frac{ln(ln(x))}{x}$$
$$Longrightarrow y' = frac{(ln(x))^{ln(x)}}{x}(1+ln(ln(x)))$$



which is the required answer.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    I think that this question is solved by this method : $u(x)^{v(x)}= u(x)^{v(x)}. [ln(u(x)).v(x)]'$ . Then,



    $ln(x)^{ln(x)}.[ln(lnx).ln(x)]'$.






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      If $y = ln(x)^{ln(x)}$, then $ln(y) = lnleft(ln(x)^{ln(x)}right) = ln(x) cdot ln(ln(x))$. The right hand side is straightforward to differentiate, using chain and product rules. I'll leave that to you.



      The left side can be differentiated with the chain rule, remembering that $y$ is a function of $x$. We do the usual: differentiate the outer function $ln$, leaving the inner function $y$ in tact, then multiply by the derivative of the inner function. That is,
      $$frac{mathrm{d}}{mathrm{d}x} ln(y) = frac{1}{y} cdot y' = frac{y'}{ln(x)^{ln(x)}}$$
      So, differentiating both sides yields
      $$frac{y'}{ln(x)^{ln(x)}} = frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)),$$
      hence
      $$y' = ln(x)^{ln(x)} cdot frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)).$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I now understand how logarithmic differentiation works after looking at your example, thanks!
        $endgroup$
        – Kyky
        Feb 1 at 9:33
















      3












      $begingroup$

      If $y = ln(x)^{ln(x)}$, then $ln(y) = lnleft(ln(x)^{ln(x)}right) = ln(x) cdot ln(ln(x))$. The right hand side is straightforward to differentiate, using chain and product rules. I'll leave that to you.



      The left side can be differentiated with the chain rule, remembering that $y$ is a function of $x$. We do the usual: differentiate the outer function $ln$, leaving the inner function $y$ in tact, then multiply by the derivative of the inner function. That is,
      $$frac{mathrm{d}}{mathrm{d}x} ln(y) = frac{1}{y} cdot y' = frac{y'}{ln(x)^{ln(x)}}$$
      So, differentiating both sides yields
      $$frac{y'}{ln(x)^{ln(x)}} = frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)),$$
      hence
      $$y' = ln(x)^{ln(x)} cdot frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)).$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I now understand how logarithmic differentiation works after looking at your example, thanks!
        $endgroup$
        – Kyky
        Feb 1 at 9:33














      3












      3








      3





      $begingroup$

      If $y = ln(x)^{ln(x)}$, then $ln(y) = lnleft(ln(x)^{ln(x)}right) = ln(x) cdot ln(ln(x))$. The right hand side is straightforward to differentiate, using chain and product rules. I'll leave that to you.



      The left side can be differentiated with the chain rule, remembering that $y$ is a function of $x$. We do the usual: differentiate the outer function $ln$, leaving the inner function $y$ in tact, then multiply by the derivative of the inner function. That is,
      $$frac{mathrm{d}}{mathrm{d}x} ln(y) = frac{1}{y} cdot y' = frac{y'}{ln(x)^{ln(x)}}$$
      So, differentiating both sides yields
      $$frac{y'}{ln(x)^{ln(x)}} = frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)),$$
      hence
      $$y' = ln(x)^{ln(x)} cdot frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)).$$






      share|cite|improve this answer









      $endgroup$



      If $y = ln(x)^{ln(x)}$, then $ln(y) = lnleft(ln(x)^{ln(x)}right) = ln(x) cdot ln(ln(x))$. The right hand side is straightforward to differentiate, using chain and product rules. I'll leave that to you.



      The left side can be differentiated with the chain rule, remembering that $y$ is a function of $x$. We do the usual: differentiate the outer function $ln$, leaving the inner function $y$ in tact, then multiply by the derivative of the inner function. That is,
      $$frac{mathrm{d}}{mathrm{d}x} ln(y) = frac{1}{y} cdot y' = frac{y'}{ln(x)^{ln(x)}}$$
      So, differentiating both sides yields
      $$frac{y'}{ln(x)^{ln(x)}} = frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)),$$
      hence
      $$y' = ln(x)^{ln(x)} cdot frac{mathrm{d}}{mathrm{d}x} (ln(x) cdot ln(ln(x)).$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Feb 1 at 8:54









      Theo BenditTheo Bendit

      20.8k12355




      20.8k12355












      • $begingroup$
        I now understand how logarithmic differentiation works after looking at your example, thanks!
        $endgroup$
        – Kyky
        Feb 1 at 9:33


















      • $begingroup$
        I now understand how logarithmic differentiation works after looking at your example, thanks!
        $endgroup$
        – Kyky
        Feb 1 at 9:33
















      $begingroup$
      I now understand how logarithmic differentiation works after looking at your example, thanks!
      $endgroup$
      – Kyky
      Feb 1 at 9:33




      $begingroup$
      I now understand how logarithmic differentiation works after looking at your example, thanks!
      $endgroup$
      – Kyky
      Feb 1 at 9:33











      2












      $begingroup$

      $$y=(ln(x))^{ln(x)}$$
      $$Longrightarrow ln(y)=ln(x)ln(ln(x))$$
      $$Longrightarrow frac{y'}{y}=frac{ln(x)}{xln(x)}+frac{ln(ln(x))}{x}$$
      $$Longrightarrow y' = frac{(ln(x))^{ln(x)}}{x}(1+ln(ln(x)))$$



      which is the required answer.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        $$y=(ln(x))^{ln(x)}$$
        $$Longrightarrow ln(y)=ln(x)ln(ln(x))$$
        $$Longrightarrow frac{y'}{y}=frac{ln(x)}{xln(x)}+frac{ln(ln(x))}{x}$$
        $$Longrightarrow y' = frac{(ln(x))^{ln(x)}}{x}(1+ln(ln(x)))$$



        which is the required answer.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          $$y=(ln(x))^{ln(x)}$$
          $$Longrightarrow ln(y)=ln(x)ln(ln(x))$$
          $$Longrightarrow frac{y'}{y}=frac{ln(x)}{xln(x)}+frac{ln(ln(x))}{x}$$
          $$Longrightarrow y' = frac{(ln(x))^{ln(x)}}{x}(1+ln(ln(x)))$$



          which is the required answer.






          share|cite|improve this answer











          $endgroup$



          $$y=(ln(x))^{ln(x)}$$
          $$Longrightarrow ln(y)=ln(x)ln(ln(x))$$
          $$Longrightarrow frac{y'}{y}=frac{ln(x)}{xln(x)}+frac{ln(ln(x))}{x}$$
          $$Longrightarrow y' = frac{(ln(x))^{ln(x)}}{x}(1+ln(ln(x)))$$



          which is the required answer.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 1 at 8:58









          Paras Khosla

          3,101626




          3,101626










          answered Feb 1 at 8:54









          MartundMartund

          1,965213




          1,965213























              0












              $begingroup$

              I think that this question is solved by this method : $u(x)^{v(x)}= u(x)^{v(x)}. [ln(u(x)).v(x)]'$ . Then,



              $ln(x)^{ln(x)}.[ln(lnx).ln(x)]'$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I think that this question is solved by this method : $u(x)^{v(x)}= u(x)^{v(x)}. [ln(u(x)).v(x)]'$ . Then,



                $ln(x)^{ln(x)}.[ln(lnx).ln(x)]'$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I think that this question is solved by this method : $u(x)^{v(x)}= u(x)^{v(x)}. [ln(u(x)).v(x)]'$ . Then,



                  $ln(x)^{ln(x)}.[ln(lnx).ln(x)]'$.






                  share|cite|improve this answer









                  $endgroup$



                  I think that this question is solved by this method : $u(x)^{v(x)}= u(x)^{v(x)}. [ln(u(x)).v(x)]'$ . Then,



                  $ln(x)^{ln(x)}.[ln(lnx).ln(x)]'$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 1 at 9:05









                  mathsstudentmathsstudent

                  536




                  536






























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