Derivative of x^(2x)












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Maybe a simple question but I can't find a good example of it on the internet.
What is the derivative of:



$x^{2x}$



What is the simplest method of determining the derivative and what kind of rules are involved?










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    0












    $begingroup$


    Maybe a simple question but I can't find a good example of it on the internet.
    What is the derivative of:



    $x^{2x}$



    What is the simplest method of determining the derivative and what kind of rules are involved?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Maybe a simple question but I can't find a good example of it on the internet.
      What is the derivative of:



      $x^{2x}$



      What is the simplest method of determining the derivative and what kind of rules are involved?










      share|cite|improve this question









      $endgroup$




      Maybe a simple question but I can't find a good example of it on the internet.
      What is the derivative of:



      $x^{2x}$



      What is the simplest method of determining the derivative and what kind of rules are involved?







      derivatives






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Oct 28 '14 at 20:47









      BasBas

      1233




      1233






















          3 Answers
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          $begingroup$

          Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..






            share|cite|improve this answer











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              $begingroup$

              As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,



              $$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.






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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                6












                $begingroup$

                Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.






                share|cite|improve this answer









                $endgroup$


















                  6












                  $begingroup$

                  Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.






                  share|cite|improve this answer









                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.






                    share|cite|improve this answer









                    $endgroup$



                    Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Oct 28 '14 at 20:48









                    TomTom

                    9,0571218




                    9,0571218























                        0












                        $begingroup$

                        HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..






                            share|cite|improve this answer











                            $endgroup$



                            HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            answered Oct 28 '14 at 20:58


























                            community wiki





                            Argha
























                                0












                                $begingroup$

                                As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,



                                $$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,



                                  $$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,



                                    $$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,



                                    $$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 2 at 3:09









                                    Axion004Axion004

                                    430413




                                    430413






























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