Are these two equivalent












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When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
$(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$






calculus







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    0












    $begingroup$


    When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
    Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
    $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$






    calculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
      Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
      $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$






      calculus







      share|cite|improve this question









      $endgroup$




      When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
      Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
      $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$






      calculus




      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 7:33









      mingming

      3365




      3365






















          2 Answers
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          $begingroup$

          They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.






          share|cite|improve this answer









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          • $begingroup$
            Why doesn't the power rule apply here as well?
            $endgroup$
            – ming
            Dec 10 '18 at 17:25










          • $begingroup$
            Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 10 '18 at 17:29





















          1












          $begingroup$

          Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !



          We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
            $endgroup$
            – ming
            Dec 10 '18 at 17:58











          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why doesn't the power rule apply here as well?
            $endgroup$
            – ming
            Dec 10 '18 at 17:25










          • $begingroup$
            Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 10 '18 at 17:29


















          1












          $begingroup$

          They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why doesn't the power rule apply here as well?
            $endgroup$
            – ming
            Dec 10 '18 at 17:25










          • $begingroup$
            Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 10 '18 at 17:29
















          1












          1








          1





          $begingroup$

          They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.






          share|cite|improve this answer









          $endgroup$



          They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 7:36









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          74.3k42865




          74.3k42865












          • $begingroup$
            Why doesn't the power rule apply here as well?
            $endgroup$
            – ming
            Dec 10 '18 at 17:25










          • $begingroup$
            Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 10 '18 at 17:29




















          • $begingroup$
            Why doesn't the power rule apply here as well?
            $endgroup$
            – ming
            Dec 10 '18 at 17:25










          • $begingroup$
            Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 10 '18 at 17:29


















          $begingroup$
          Why doesn't the power rule apply here as well?
          $endgroup$
          – ming
          Dec 10 '18 at 17:25




          $begingroup$
          Why doesn't the power rule apply here as well?
          $endgroup$
          – ming
          Dec 10 '18 at 17:25












          $begingroup$
          Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
          $endgroup$
          – Dr. Sonnhard Graubner
          Dec 10 '18 at 17:29






          $begingroup$
          Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
          $endgroup$
          – Dr. Sonnhard Graubner
          Dec 10 '18 at 17:29













          1












          $begingroup$

          Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !



          We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
            $endgroup$
            – ming
            Dec 10 '18 at 17:58
















          1












          $begingroup$

          Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !



          We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
            $endgroup$
            – ming
            Dec 10 '18 at 17:58














          1












          1








          1





          $begingroup$

          Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !



          We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$






          share|cite|improve this answer









          $endgroup$



          Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !



          We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 7:36









          FredFred

          45.1k1847




          45.1k1847












          • $begingroup$
            So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
            $endgroup$
            – ming
            Dec 10 '18 at 17:58


















          • $begingroup$
            So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
            $endgroup$
            – ming
            Dec 10 '18 at 17:58
















          $begingroup$
          So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
          $endgroup$
          – ming
          Dec 10 '18 at 17:58




          $begingroup$
          So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
          $endgroup$
          – ming
          Dec 10 '18 at 17:58


















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