Confused about the difference between these two derivatives












1












$begingroup$


So for a function $f(x) = e^{sin(x^2)}$, you're supposed to apply the chain rule to it to get $e^{sin(x^2)}(cos(x^2))(2x)$.



But for a function $f(x) = 5^{cos(x)}$, you apply the $a^x(ln(a))$ rule, to get $5^{cos(x)}ln(5)(-sin(x))$??



I'm confused, cause $5$ and $e$ are both numbers with a trig function as its exponent, why do we treat them differently?






calculus







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$endgroup$








  • 1




    $begingroup$
    Do you know what $ln(e)$ is?
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 19:59










  • $begingroup$
    It's 1, correct?
    $endgroup$
    – ming
    Dec 10 '18 at 20:33
















1












$begingroup$


So for a function $f(x) = e^{sin(x^2)}$, you're supposed to apply the chain rule to it to get $e^{sin(x^2)}(cos(x^2))(2x)$.



But for a function $f(x) = 5^{cos(x)}$, you apply the $a^x(ln(a))$ rule, to get $5^{cos(x)}ln(5)(-sin(x))$??



I'm confused, cause $5$ and $e$ are both numbers with a trig function as its exponent, why do we treat them differently?






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you know what $ln(e)$ is?
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 19:59










  • $begingroup$
    It's 1, correct?
    $endgroup$
    – ming
    Dec 10 '18 at 20:33














1












1








1





$begingroup$


So for a function $f(x) = e^{sin(x^2)}$, you're supposed to apply the chain rule to it to get $e^{sin(x^2)}(cos(x^2))(2x)$.



But for a function $f(x) = 5^{cos(x)}$, you apply the $a^x(ln(a))$ rule, to get $5^{cos(x)}ln(5)(-sin(x))$??



I'm confused, cause $5$ and $e$ are both numbers with a trig function as its exponent, why do we treat them differently?






calculus







share|cite|improve this question











$endgroup$




So for a function $f(x) = e^{sin(x^2)}$, you're supposed to apply the chain rule to it to get $e^{sin(x^2)}(cos(x^2))(2x)$.



But for a function $f(x) = 5^{cos(x)}$, you apply the $a^x(ln(a))$ rule, to get $5^{cos(x)}ln(5)(-sin(x))$??



I'm confused, cause $5$ and $e$ are both numbers with a trig function as its exponent, why do we treat them differently?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 20:00









Bernard

121k740116




121k740116










asked Dec 10 '18 at 19:57









mingming

3435




3435








  • 1




    $begingroup$
    Do you know what $ln(e)$ is?
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 19:59










  • $begingroup$
    It's 1, correct?
    $endgroup$
    – ming
    Dec 10 '18 at 20:33














  • 1




    $begingroup$
    Do you know what $ln(e)$ is?
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 19:59










  • $begingroup$
    It's 1, correct?
    $endgroup$
    – ming
    Dec 10 '18 at 20:33








1




1




$begingroup$
Do you know what $ln(e)$ is?
$endgroup$
– T. Bongers
Dec 10 '18 at 19:59




$begingroup$
Do you know what $ln(e)$ is?
$endgroup$
– T. Bongers
Dec 10 '18 at 19:59












$begingroup$
It's 1, correct?
$endgroup$
– ming
Dec 10 '18 at 20:33




$begingroup$
It's 1, correct?
$endgroup$
– ming
Dec 10 '18 at 20:33










2 Answers
2






active

oldest

votes


















3












$begingroup$

You don't treat them differently. In fact, you have
$$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
$$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
since $ln (e) = 1$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    For $a>0$ and $xin Bbb R$,



    $$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$



    the derivative is



    $$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      3












      $begingroup$

      You don't treat them differently. In fact, you have
      $$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
      for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
      $$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
      since $ln (e) = 1$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        You don't treat them differently. In fact, you have
        $$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
        for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
        $$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
        since $ln (e) = 1$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          You don't treat them differently. In fact, you have
          $$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
          for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
          $$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
          since $ln (e) = 1$.






          share|cite|improve this answer











          $endgroup$



          You don't treat them differently. In fact, you have
          $$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
          for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
          $$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
          since $ln (e) = 1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 20:06









          Chickenmancer

          3,314724




          3,314724










          answered Dec 10 '18 at 19:59









          MisterRiemannMisterRiemann

          5,8121625




          5,8121625























              0












              $begingroup$

              For $a>0$ and $xin Bbb R$,



              $$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$



              the derivative is



              $$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                For $a>0$ and $xin Bbb R$,



                $$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$



                the derivative is



                $$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  For $a>0$ and $xin Bbb R$,



                  $$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$



                  the derivative is



                  $$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$






                  share|cite|improve this answer









                  $endgroup$



                  For $a>0$ and $xin Bbb R$,



                  $$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$



                  the derivative is



                  $$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 20:02









                  hamam_Abdallahhamam_Abdallah

                  38k21634




                  38k21634






























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