Why can't I use the formula for $f(x)=x^x$ to derive $(frac{x}{x+1})^x$?












0












$begingroup$


Why can't I use the formula $y=x^x$, $dot {y}=x^x(ln x+1)$ to derive $(frac{x}{x+1})^x$?
I've tried and it doesn't get me the correct result. Have I done any mistake?
$$Bigl[Bigl(frac{x}{x+1}Bigr)^xBigr]'=(frac{x}{x+1})^xBigl[lnBigl(frac{x}{x+1}Bigr)+1Bigr]Bigl[frac{1+x-x}{(x+1)^2}Bigr]=frac{bigl(frac{x}{x+1}bigr)^xBigl[lnbigl(frac{x}{x+1}bigr)+1Bigr]}{(1+x)^2}$$
If not, please explain why I can't use the formula.






derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Simple: you need the quotient and chain rules.
    $endgroup$
    – Sean Roberson
    Jan 10 at 19:10
















0












$begingroup$


Why can't I use the formula $y=x^x$, $dot {y}=x^x(ln x+1)$ to derive $(frac{x}{x+1})^x$?
I've tried and it doesn't get me the correct result. Have I done any mistake?
$$Bigl[Bigl(frac{x}{x+1}Bigr)^xBigr]'=(frac{x}{x+1})^xBigl[lnBigl(frac{x}{x+1}Bigr)+1Bigr]Bigl[frac{1+x-x}{(x+1)^2}Bigr]=frac{bigl(frac{x}{x+1}bigr)^xBigl[lnbigl(frac{x}{x+1}bigr)+1Bigr]}{(1+x)^2}$$
If not, please explain why I can't use the formula.






derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Simple: you need the quotient and chain rules.
    $endgroup$
    – Sean Roberson
    Jan 10 at 19:10














0












0








0





$begingroup$


Why can't I use the formula $y=x^x$, $dot {y}=x^x(ln x+1)$ to derive $(frac{x}{x+1})^x$?
I've tried and it doesn't get me the correct result. Have I done any mistake?
$$Bigl[Bigl(frac{x}{x+1}Bigr)^xBigr]'=(frac{x}{x+1})^xBigl[lnBigl(frac{x}{x+1}Bigr)+1Bigr]Bigl[frac{1+x-x}{(x+1)^2}Bigr]=frac{bigl(frac{x}{x+1}bigr)^xBigl[lnbigl(frac{x}{x+1}bigr)+1Bigr]}{(1+x)^2}$$
If not, please explain why I can't use the formula.






derivatives







share|cite|improve this question











$endgroup$




Why can't I use the formula $y=x^x$, $dot {y}=x^x(ln x+1)$ to derive $(frac{x}{x+1})^x$?
I've tried and it doesn't get me the correct result. Have I done any mistake?
$$Bigl[Bigl(frac{x}{x+1}Bigr)^xBigr]'=(frac{x}{x+1})^xBigl[lnBigl(frac{x}{x+1}Bigr)+1Bigr]Bigl[frac{1+x-x}{(x+1)^2}Bigr]=frac{bigl(frac{x}{x+1}bigr)^xBigl[lnbigl(frac{x}{x+1}bigr)+1Bigr]}{(1+x)^2}$$
If not, please explain why I can't use the formula.






derivatives




derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 19:14









Bernard

122k740116




122k740116










asked Jan 10 at 19:10









Lauren SinLauren Sin

1055




1055








  • 2




    $begingroup$
    Simple: you need the quotient and chain rules.
    $endgroup$
    – Sean Roberson
    Jan 10 at 19:10














  • 2




    $begingroup$
    Simple: you need the quotient and chain rules.
    $endgroup$
    – Sean Roberson
    Jan 10 at 19:10








2




2




$begingroup$
Simple: you need the quotient and chain rules.
$endgroup$
– Sean Roberson
Jan 10 at 19:10




$begingroup$
Simple: you need the quotient and chain rules.
$endgroup$
– Sean Roberson
Jan 10 at 19:10










5 Answers
5






active

oldest

votes


















2












$begingroup$

Hint:



Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
$$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
+x,frac{u'(x)}{u(x)}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
    $endgroup$
    – Bernard
    Jan 10 at 20:40



















2












$begingroup$

Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
      left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
      right) +1 right) }
      $$
      Taking the logarithm on both sides we get
      $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
      $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$






      share|cite|improve this answer











      $endgroup$





















        1












        $begingroup$

        What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$






        share|cite|improve this answer









        $endgroup$













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069062%2fwhy-cant-i-use-the-formula-for-fx-xx-to-derive-fracxx1x%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Hint:



          Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
          $$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
          +x,frac{u'(x)}{u(x)}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
            $endgroup$
            – Bernard
            Jan 10 at 20:40
















          2












          $begingroup$

          Hint:



          Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
          $$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
          +x,frac{u'(x)}{u(x)}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
            $endgroup$
            – Bernard
            Jan 10 at 20:40














          2












          2








          2





          $begingroup$

          Hint:



          Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
          $$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
          +x,frac{u'(x)}{u(x)}.$$






          share|cite|improve this answer











          $endgroup$



          Hint:



          Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
          $$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
          +x,frac{u'(x)}{u(x)}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 20:39

























          answered Jan 10 at 19:32









          BernardBernard

          122k740116




          122k740116












          • $begingroup$
            Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
            $endgroup$
            – Bernard
            Jan 10 at 20:40


















          • $begingroup$
            Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
            $endgroup$
            – Bernard
            Jan 10 at 20:40
















          $begingroup$
          Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
          $endgroup$
          – Bernard
          Jan 10 at 20:40




          $begingroup$
          Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
          $endgroup$
          – Bernard
          Jan 10 at 20:40











          2












          $begingroup$

          Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.






              share|cite|improve this answer









              $endgroup$



              Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 10 at 19:16









              J.G.J.G.

              28.7k22845




              28.7k22845























                  2












                  $begingroup$

                  To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.






                      share|cite|improve this answer









                      $endgroup$



                      To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 10 at 19:22









                      Martin ArgeramiMartin Argerami

                      128k1183183




                      128k1183183























                          1












                          $begingroup$

                          Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
                          left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
                          right) +1 right) }
                          $$
                          Taking the logarithm on both sides we get
                          $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
                          $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$






                          share|cite|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
                            left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
                            right) +1 right) }
                            $$
                            Taking the logarithm on both sides we get
                            $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
                            $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$






                            share|cite|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
                              left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
                              right) +1 right) }
                              $$
                              Taking the logarithm on both sides we get
                              $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
                              $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$






                              share|cite|improve this answer











                              $endgroup$



                              Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
                              left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
                              right) +1 right) }
                              $$
                              Taking the logarithm on both sides we get
                              $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
                              $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jan 10 at 19:34

























                              answered Jan 10 at 19:14









                              Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                              76.8k42866




                              76.8k42866























                                  1












                                  $begingroup$

                                  What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 10 at 20:19









                                      Mostafa AyazMostafa Ayaz

                                      15.7k3939




                                      15.7k3939






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069062%2fwhy-cant-i-use-the-formula-for-fx-xx-to-derive-fracxx1x%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          Human spaceflight

                                          Image
                                          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
                                          Read more

                                          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                          Image
                                          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
                                          Read more

                                          File:DeusFollowingSea.jpg

                                          Image
                                          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
                                          Read more