Why can't I differentiate $ x^{sin x} $ using the power rule?












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$begingroup$


I'm trying to differentiate $ x^{sin x} $, with respect to $ x $ and $x > 0 $. My textbook initiates with $ y = x^{sin x} $, takes logarithms on both sides and arrives at the answer $$ x^{sin x - 1}.sin x + x^{sin x}.cos x log x $$



Why can't I use the power rule to proceed like this: $ y' = (sin x) x^{sin x-1} cos x $.



Here, I've first differentiated $ x $ with respect to $ sin x $ and then I've differentiated $ sin x $ with respect to $ x $ to get $ cos x $.






derivatives







share|cite|improve this question









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    4












    $begingroup$


    I'm trying to differentiate $ x^{sin x} $, with respect to $ x $ and $x > 0 $. My textbook initiates with $ y = x^{sin x} $, takes logarithms on both sides and arrives at the answer $$ x^{sin x - 1}.sin x + x^{sin x}.cos x log x $$



    Why can't I use the power rule to proceed like this: $ y' = (sin x) x^{sin x-1} cos x $.



    Here, I've first differentiated $ x $ with respect to $ sin x $ and then I've differentiated $ sin x $ with respect to $ x $ to get $ cos x $.






    derivatives







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      I'm trying to differentiate $ x^{sin x} $, with respect to $ x $ and $x > 0 $. My textbook initiates with $ y = x^{sin x} $, takes logarithms on both sides and arrives at the answer $$ x^{sin x - 1}.sin x + x^{sin x}.cos x log x $$



      Why can't I use the power rule to proceed like this: $ y' = (sin x) x^{sin x-1} cos x $.



      Here, I've first differentiated $ x $ with respect to $ sin x $ and then I've differentiated $ sin x $ with respect to $ x $ to get $ cos x $.






      derivatives







      share|cite|improve this question









      $endgroup$




      I'm trying to differentiate $ x^{sin x} $, with respect to $ x $ and $x > 0 $. My textbook initiates with $ y = x^{sin x} $, takes logarithms on both sides and arrives at the answer $$ x^{sin x - 1}.sin x + x^{sin x}.cos x log x $$



      Why can't I use the power rule to proceed like this: $ y' = (sin x) x^{sin x-1} cos x $.



      Here, I've first differentiated $ x $ with respect to $ sin x $ and then I've differentiated $ sin x $ with respect to $ x $ to get $ cos x $.






      derivatives




      derivatives






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      asked Dec 25 '18 at 11:29









      WorldGovWorldGov

      345211




      345211






















          4 Answers
          4






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          oldest

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          4












          $begingroup$

          The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.




          To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does that proof hold for $k lt 0$?
            $endgroup$
            – user150203
            Jan 6 at 11:44












          • $begingroup$
            Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
            $endgroup$
            – user150203
            Jan 6 at 11:48










          • $begingroup$
            No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
            $endgroup$
            – user150203
            Jan 6 at 11:50










          • $begingroup$
            I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
            $endgroup$
            – user150203
            Jan 6 at 11:53












          • $begingroup$
            @DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
            $endgroup$
            – TheSimpliFire
            Jan 6 at 11:54





















          1












          $begingroup$

          $(x^k)'=kx^{k-1},$ where $k$ is constant.



          In our case $sin{x}neq constant$ and we can not use this rule.



          By the way,
          $$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
          Can you end it now?






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              You are trying to use the rule
              $$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$



              But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
                $endgroup$
                – Michael Rozenberg
                Dec 25 '18 at 11:49












              Your Answer





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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4












              $begingroup$

              The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.




              To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Does that proof hold for $k lt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:44












              • $begingroup$
                Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
                $endgroup$
                – user150203
                Jan 6 at 11:48










              • $begingroup$
                No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:50










              • $begingroup$
                I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:53












              • $begingroup$
                @DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
                $endgroup$
                – TheSimpliFire
                Jan 6 at 11:54


















              4












              $begingroup$

              The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.




              To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Does that proof hold for $k lt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:44












              • $begingroup$
                Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
                $endgroup$
                – user150203
                Jan 6 at 11:48










              • $begingroup$
                No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:50










              • $begingroup$
                I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:53












              • $begingroup$
                @DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
                $endgroup$
                – TheSimpliFire
                Jan 6 at 11:54
















              4












              4








              4





              $begingroup$

              The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.




              To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.






              share|cite|improve this answer









              $endgroup$



              The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.




              To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 25 '18 at 11:30









              TheSimpliFireTheSimpliFire

              13.1k62464




              13.1k62464












              • $begingroup$
                Does that proof hold for $k lt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:44












              • $begingroup$
                Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
                $endgroup$
                – user150203
                Jan 6 at 11:48










              • $begingroup$
                No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:50










              • $begingroup$
                I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:53












              • $begingroup$
                @DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
                $endgroup$
                – TheSimpliFire
                Jan 6 at 11:54




















              • $begingroup$
                Does that proof hold for $k lt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:44












              • $begingroup$
                Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
                $endgroup$
                – user150203
                Jan 6 at 11:48










              • $begingroup$
                No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:50










              • $begingroup$
                I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
                $endgroup$
                – user150203
                Jan 6 at 11:53












              • $begingroup$
                @DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
                $endgroup$
                – TheSimpliFire
                Jan 6 at 11:54


















              $begingroup$
              Does that proof hold for $k lt 0$?
              $endgroup$
              – user150203
              Jan 6 at 11:44






              $begingroup$
              Does that proof hold for $k lt 0$?
              $endgroup$
              – user150203
              Jan 6 at 11:44














              $begingroup$
              Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
              $endgroup$
              – user150203
              Jan 6 at 11:48




              $begingroup$
              Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
              $endgroup$
              – user150203
              Jan 6 at 11:48












              $begingroup$
              No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
              $endgroup$
              – user150203
              Jan 6 at 11:50




              $begingroup$
              No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
              $endgroup$
              – user150203
              Jan 6 at 11:50












              $begingroup$
              I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
              $endgroup$
              – user150203
              Jan 6 at 11:53






              $begingroup$
              I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
              $endgroup$
              – user150203
              Jan 6 at 11:53














              $begingroup$
              @DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
              $endgroup$
              – TheSimpliFire
              Jan 6 at 11:54






              $begingroup$
              @DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
              $endgroup$
              – TheSimpliFire
              Jan 6 at 11:54













              1












              $begingroup$

              $(x^k)'=kx^{k-1},$ where $k$ is constant.



              In our case $sin{x}neq constant$ and we can not use this rule.



              By the way,
              $$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
              Can you end it now?






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                $(x^k)'=kx^{k-1},$ where $k$ is constant.



                In our case $sin{x}neq constant$ and we can not use this rule.



                By the way,
                $$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
                Can you end it now?






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $(x^k)'=kx^{k-1},$ where $k$ is constant.



                  In our case $sin{x}neq constant$ and we can not use this rule.



                  By the way,
                  $$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
                  Can you end it now?






                  share|cite|improve this answer











                  $endgroup$



                  $(x^k)'=kx^{k-1},$ where $k$ is constant.



                  In our case $sin{x}neq constant$ and we can not use this rule.



                  By the way,
                  $$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
                  Can you end it now?







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 25 '18 at 11:45

























                  answered Dec 25 '18 at 11:38









                  Michael RozenbergMichael Rozenberg

                  110k1896201




                  110k1896201























                      1












                      $begingroup$

                      Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.






                          share|cite|improve this answer









                          $endgroup$



                          Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 25 '18 at 11:55









                          The_SympathizerThe_Sympathizer

                          7,8002246




                          7,8002246























                              0












                              $begingroup$

                              You are trying to use the rule
                              $$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$



                              But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
                                $endgroup$
                                – Michael Rozenberg
                                Dec 25 '18 at 11:49
















                              0












                              $begingroup$

                              You are trying to use the rule
                              $$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$



                              But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
                                $endgroup$
                                – Michael Rozenberg
                                Dec 25 '18 at 11:49














                              0












                              0








                              0





                              $begingroup$

                              You are trying to use the rule
                              $$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$



                              But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.






                              share|cite|improve this answer









                              $endgroup$



                              You are trying to use the rule
                              $$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$



                              But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 25 '18 at 11:39









                              TonyKTonyK

                              43.9k358137




                              43.9k358137












                              • $begingroup$
                                Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
                                $endgroup$
                                – Michael Rozenberg
                                Dec 25 '18 at 11:49


















                              • $begingroup$
                                Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
                                $endgroup$
                                – Michael Rozenberg
                                Dec 25 '18 at 11:49
















                              $begingroup$
                              Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
                              $endgroup$
                              – Michael Rozenberg
                              Dec 25 '18 at 11:49




                              $begingroup$
                              Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
                              $endgroup$
                              – Michael Rozenberg
                              Dec 25 '18 at 11:49


















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                              Verónica Boquete

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                              Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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                              Ida-Boy-Ed-Garten

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                              Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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