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









$endgroup$

















    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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      share|cite|improve this question











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      asked Dec 25 '18 at 11:29









      WorldGovWorldGov

      324111




      324111






















          4 Answers
          4






          active

          oldest

          votes


















          3












          $begingroup$

          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 $k lt 0$?
            $endgroup$
            – DavidG
            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 $x gt 0$ ?
            $endgroup$
            – DavidG
            Jan 6 at 11:48










          • $begingroup$
            No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
            $endgroup$
            – DavidG
            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 $x gt 0$?
            $endgroup$
            – DavidG
            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











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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              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 $k lt 0$?
                $endgroup$
                – DavidG
                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 $x gt 0$ ?
                $endgroup$
                – DavidG
                Jan 6 at 11:48










              • $begingroup$
                No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
                $endgroup$
                – DavidG
                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 $x gt 0$?
                $endgroup$
                – DavidG
                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


















              3












              $begingroup$

              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 $k lt 0$?
                $endgroup$
                – DavidG
                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 $x gt 0$ ?
                $endgroup$
                – DavidG
                Jan 6 at 11:48










              • $begingroup$
                No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
                $endgroup$
                – DavidG
                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 $x gt 0$?
                $endgroup$
                – DavidG
                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
















              3












              3








              3





              $begingroup$

              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

              12.7k62461




              12.7k62461












              • $begingroup$
                Does that proof hold for $k lt 0$?
                $endgroup$
                – DavidG
                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 $x gt 0$ ?
                $endgroup$
                – DavidG
                Jan 6 at 11:48










              • $begingroup$
                No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
                $endgroup$
                – DavidG
                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 $x gt 0$?
                $endgroup$
                – DavidG
                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 $k lt 0$?
                $endgroup$
                – DavidG
                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 $x gt 0$ ?
                $endgroup$
                – DavidG
                Jan 6 at 11:48










              • $begingroup$
                No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
                $endgroup$
                – DavidG
                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 $x gt 0$?
                $endgroup$
                – DavidG
                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 $k lt 0$?
              $endgroup$
              – DavidG
              Jan 6 at 11:44






              $begingroup$
              Does that proof hold for $k lt 0$?
              $endgroup$
              – DavidG
              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 $x gt 0$ ?
              $endgroup$
              – DavidG
              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 $x gt 0$ ?
              $endgroup$
              – DavidG
              Jan 6 at 11:48












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




              $begingroup$
              No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
              $endgroup$
              – DavidG
              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 $x gt 0$?
              $endgroup$
              – DavidG
              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 $x gt 0$?
              $endgroup$
              – DavidG
              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

                  107k1895199




                  107k1895199























                      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

                          7,7402246




                          7,7402246























                              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

                              43k356135




                              43k356135












                              • $begingroup$
                                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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                              Marschland

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                              Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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                              Dieringhausen

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                              Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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