Finding Maclaurin’s series expansion function of $f(x) = a^x$ at $x=0$












1












$begingroup$


First derivation is easy as it's $a^x * log a$, but I have some troubles with finding the derivative of $ a^x * log a$



Using the product rule I have $ (a^x *log a)' = (a^x * log^2 a) + a^x * frac{1}{a}$, whereas wolfram says it should be just $a^x * log^2 a$.



Need a hint.






calculus derivatives taylor-expansion







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  • 1




    $begingroup$
    The answer below solves your problem with the 2nd differentiation, and for the whole work perhaps you may want to consider $;a^x=e^{xlog a};$
    $endgroup$
    – DonAntonio
    Dec 10 '18 at 10:03
















1












$begingroup$


First derivation is easy as it's $a^x * log a$, but I have some troubles with finding the derivative of $ a^x * log a$



Using the product rule I have $ (a^x *log a)' = (a^x * log^2 a) + a^x * frac{1}{a}$, whereas wolfram says it should be just $a^x * log^2 a$.



Need a hint.






calculus derivatives taylor-expansion







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The answer below solves your problem with the 2nd differentiation, and for the whole work perhaps you may want to consider $;a^x=e^{xlog a};$
    $endgroup$
    – DonAntonio
    Dec 10 '18 at 10:03














1












1








1


0



$begingroup$


First derivation is easy as it's $a^x * log a$, but I have some troubles with finding the derivative of $ a^x * log a$



Using the product rule I have $ (a^x *log a)' = (a^x * log^2 a) + a^x * frac{1}{a}$, whereas wolfram says it should be just $a^x * log^2 a$.



Need a hint.






calculus derivatives taylor-expansion







share|cite|improve this question









$endgroup$




First derivation is easy as it's $a^x * log a$, but I have some troubles with finding the derivative of $ a^x * log a$



Using the product rule I have $ (a^x *log a)' = (a^x * log^2 a) + a^x * frac{1}{a}$, whereas wolfram says it should be just $a^x * log^2 a$.



Need a hint.






calculus derivatives taylor-expansion




calculus derivatives taylor-expansion






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asked Dec 10 '18 at 10:00









BartoszBartosz

285




285








  • 1




    $begingroup$
    The answer below solves your problem with the 2nd differentiation, and for the whole work perhaps you may want to consider $;a^x=e^{xlog a};$
    $endgroup$
    – DonAntonio
    Dec 10 '18 at 10:03














  • 1




    $begingroup$
    The answer below solves your problem with the 2nd differentiation, and for the whole work perhaps you may want to consider $;a^x=e^{xlog a};$
    $endgroup$
    – DonAntonio
    Dec 10 '18 at 10:03








1




1




$begingroup$
The answer below solves your problem with the 2nd differentiation, and for the whole work perhaps you may want to consider $;a^x=e^{xlog a};$
$endgroup$
– DonAntonio
Dec 10 '18 at 10:03




$begingroup$
The answer below solves your problem with the 2nd differentiation, and for the whole work perhaps you may want to consider $;a^x=e^{xlog a};$
$endgroup$
– DonAntonio
Dec 10 '18 at 10:03










3 Answers
3






active

oldest

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2












$begingroup$

Hint: Differentiation is with respect to $x $.






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  • 4




    $begingroup$
    Haha, that's a pretty useful hint, thank you
    $endgroup$
    – Bartosz
    Dec 10 '18 at 10:11



















0












$begingroup$

We get $$(a^x)'=a^xlog(a)$$
$$(a^xlog(a))'=a^x(log(a))^2$$
$$(a^x(log(a))^2)'=a^x(log(a))^3$$ etc.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Note that
    $$
    a^x=e^{xlog{a}}
    $$
    You know how to take derivatives of the exponential.






    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









      2












      $begingroup$

      Hint: Differentiation is with respect to $x $.






      share|cite|improve this answer









      $endgroup$









      • 4




        $begingroup$
        Haha, that's a pretty useful hint, thank you
        $endgroup$
        – Bartosz
        Dec 10 '18 at 10:11
















      2












      $begingroup$

      Hint: Differentiation is with respect to $x $.






      share|cite|improve this answer









      $endgroup$









      • 4




        $begingroup$
        Haha, that's a pretty useful hint, thank you
        $endgroup$
        – Bartosz
        Dec 10 '18 at 10:11














      2












      2








      2





      $begingroup$

      Hint: Differentiation is with respect to $x $.






      share|cite|improve this answer









      $endgroup$



      Hint: Differentiation is with respect to $x $.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 10 '18 at 10:01









      Thomas ShelbyThomas Shelby

      2,765421




      2,765421








      • 4




        $begingroup$
        Haha, that's a pretty useful hint, thank you
        $endgroup$
        – Bartosz
        Dec 10 '18 at 10:11














      • 4




        $begingroup$
        Haha, that's a pretty useful hint, thank you
        $endgroup$
        – Bartosz
        Dec 10 '18 at 10:11








      4




      4




      $begingroup$
      Haha, that's a pretty useful hint, thank you
      $endgroup$
      – Bartosz
      Dec 10 '18 at 10:11




      $begingroup$
      Haha, that's a pretty useful hint, thank you
      $endgroup$
      – Bartosz
      Dec 10 '18 at 10:11











      0












      $begingroup$

      We get $$(a^x)'=a^xlog(a)$$
      $$(a^xlog(a))'=a^x(log(a))^2$$
      $$(a^x(log(a))^2)'=a^x(log(a))^3$$ etc.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        We get $$(a^x)'=a^xlog(a)$$
        $$(a^xlog(a))'=a^x(log(a))^2$$
        $$(a^x(log(a))^2)'=a^x(log(a))^3$$ etc.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          We get $$(a^x)'=a^xlog(a)$$
          $$(a^xlog(a))'=a^x(log(a))^2$$
          $$(a^x(log(a))^2)'=a^x(log(a))^3$$ etc.






          share|cite|improve this answer









          $endgroup$



          We get $$(a^x)'=a^xlog(a)$$
          $$(a^xlog(a))'=a^x(log(a))^2$$
          $$(a^x(log(a))^2)'=a^x(log(a))^3$$ etc.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 10:06









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          74.8k42865




          74.8k42865























              0












              $begingroup$

              Note that
              $$
              a^x=e^{xlog{a}}
              $$
              You know how to take derivatives of the exponential.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Note that
                $$
                a^x=e^{xlog{a}}
                $$
                You know how to take derivatives of the exponential.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Note that
                  $$
                  a^x=e^{xlog{a}}
                  $$
                  You know how to take derivatives of the exponential.






                  share|cite|improve this answer









                  $endgroup$



                  Note that
                  $$
                  a^x=e^{xlog{a}}
                  $$
                  You know how to take derivatives of the exponential.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 10:13









                  plus1plus1

                  3911




                  3911






























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