derivative with quotient rule and summations












1












$begingroup$


I want to differentiate this with respect to $eta$:



$$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}$$



Does this solution look correct? I tried going through with the quotient rule, but it is a bit messy, so I was hoping someone could check my math



$$C'(eta) = frac{left[3left(1-frac{eta}{eta_c}right)sumlimits_{k=1}^{3} B_kkleft(frac{eta}{eta_c}right)^{k-1}frac{1}{eta_c}right]-left[left(1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^kright)left(-frac{3}{eta_c}right)right]}{left[3left(1-frac{eta}{eta_c}right)right]^2}$$






derivatives partial-derivative







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  • 2




    $begingroup$
    Yes, that's correct.
    $endgroup$
    – Ryan Greyling
    Jan 14 at 23:09
















1












$begingroup$


I want to differentiate this with respect to $eta$:



$$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}$$



Does this solution look correct? I tried going through with the quotient rule, but it is a bit messy, so I was hoping someone could check my math



$$C'(eta) = frac{left[3left(1-frac{eta}{eta_c}right)sumlimits_{k=1}^{3} B_kkleft(frac{eta}{eta_c}right)^{k-1}frac{1}{eta_c}right]-left[left(1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^kright)left(-frac{3}{eta_c}right)right]}{left[3left(1-frac{eta}{eta_c}right)right]^2}$$






derivatives partial-derivative







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$endgroup$








  • 2




    $begingroup$
    Yes, that's correct.
    $endgroup$
    – Ryan Greyling
    Jan 14 at 23:09














1












1








1


0



$begingroup$


I want to differentiate this with respect to $eta$:



$$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}$$



Does this solution look correct? I tried going through with the quotient rule, but it is a bit messy, so I was hoping someone could check my math



$$C'(eta) = frac{left[3left(1-frac{eta}{eta_c}right)sumlimits_{k=1}^{3} B_kkleft(frac{eta}{eta_c}right)^{k-1}frac{1}{eta_c}right]-left[left(1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^kright)left(-frac{3}{eta_c}right)right]}{left[3left(1-frac{eta}{eta_c}right)right]^2}$$






derivatives partial-derivative







share|cite|improve this question









$endgroup$




I want to differentiate this with respect to $eta$:



$$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}$$



Does this solution look correct? I tried going through with the quotient rule, but it is a bit messy, so I was hoping someone could check my math



$$C'(eta) = frac{left[3left(1-frac{eta}{eta_c}right)sumlimits_{k=1}^{3} B_kkleft(frac{eta}{eta_c}right)^{k-1}frac{1}{eta_c}right]-left[left(1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^kright)left(-frac{3}{eta_c}right)right]}{left[3left(1-frac{eta}{eta_c}right)right]^2}$$






derivatives partial-derivative




derivatives partial-derivative






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asked Jan 14 at 22:57









Jackson HartJackson Hart

5452726




5452726








  • 2




    $begingroup$
    Yes, that's correct.
    $endgroup$
    – Ryan Greyling
    Jan 14 at 23:09














  • 2




    $begingroup$
    Yes, that's correct.
    $endgroup$
    – Ryan Greyling
    Jan 14 at 23:09








2




2




$begingroup$
Yes, that's correct.
$endgroup$
– Ryan Greyling
Jan 14 at 23:09




$begingroup$
Yes, that's correct.
$endgroup$
– Ryan Greyling
Jan 14 at 23:09










2 Answers
2






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0












$begingroup$

You could have made life a bit simpler writing
$$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kx^k}{3left(1-xright)}qquad text{where}qquad x=frac{eta}{eta_c}$$ and use
$$frac{dC(eta)}{deta} = frac{d C(eta)}{dx}frac{dx}{deta}=frac{1}{eta_c}frac{d C(eta)}{dx}$$ For sure, your result is correct but, since $k$ has only a few values, you could simplity the final result to get
$$C'(eta)=frac{1}{eta_c}frac{1+B_1+2 B_2 x+(3 B_3-B_2) x^2-2 B_3 x^3 }{3(1-x)^2}$$






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    0












    $begingroup$

    Here's how you could differentiate this particular function:



    $$
    C'(eta) = left[frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}right]'=
    left[frac{1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}right]'=\
    left[left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left[3left(1-frac{eta}{eta_c}right)right]^{-1}right]'=\
    left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)'left[3left(1-frac{eta}{eta_c}right)right]^{-1}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left(left[3left(1-frac{eta}{eta_c}right)right]^{-1}right)'=\
    frac{B_{1}frac{1}{eta_c}+B_{2}frac{2eta}{eta_c^2}+B_{3}frac{3eta^2}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)(-1)left[3left(1-frac{eta}{eta_c}right)right]^{-2}left(3-3frac{eta}{eta_c}right)'=\
    frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{left[3left(1-frac{eta}{eta_c}right)right]^2}cdotfrac{3}{eta_c}=\
    frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3eta_cleft(1-frac{eta}{eta_c}right)^2}.
    $$



    The answer I get appears to be equivalent to what you got.






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      2 Answers
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      2 Answers
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      0












      $begingroup$

      You could have made life a bit simpler writing
      $$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kx^k}{3left(1-xright)}qquad text{where}qquad x=frac{eta}{eta_c}$$ and use
      $$frac{dC(eta)}{deta} = frac{d C(eta)}{dx}frac{dx}{deta}=frac{1}{eta_c}frac{d C(eta)}{dx}$$ For sure, your result is correct but, since $k$ has only a few values, you could simplity the final result to get
      $$C'(eta)=frac{1}{eta_c}frac{1+B_1+2 B_2 x+(3 B_3-B_2) x^2-2 B_3 x^3 }{3(1-x)^2}$$






      share|cite|improve this answer









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        0












        $begingroup$

        You could have made life a bit simpler writing
        $$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kx^k}{3left(1-xright)}qquad text{where}qquad x=frac{eta}{eta_c}$$ and use
        $$frac{dC(eta)}{deta} = frac{d C(eta)}{dx}frac{dx}{deta}=frac{1}{eta_c}frac{d C(eta)}{dx}$$ For sure, your result is correct but, since $k$ has only a few values, you could simplity the final result to get
        $$C'(eta)=frac{1}{eta_c}frac{1+B_1+2 B_2 x+(3 B_3-B_2) x^2-2 B_3 x^3 }{3(1-x)^2}$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          You could have made life a bit simpler writing
          $$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kx^k}{3left(1-xright)}qquad text{where}qquad x=frac{eta}{eta_c}$$ and use
          $$frac{dC(eta)}{deta} = frac{d C(eta)}{dx}frac{dx}{deta}=frac{1}{eta_c}frac{d C(eta)}{dx}$$ For sure, your result is correct but, since $k$ has only a few values, you could simplity the final result to get
          $$C'(eta)=frac{1}{eta_c}frac{1+B_1+2 B_2 x+(3 B_3-B_2) x^2-2 B_3 x^3 }{3(1-x)^2}$$






          share|cite|improve this answer









          $endgroup$



          You could have made life a bit simpler writing
          $$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kx^k}{3left(1-xright)}qquad text{where}qquad x=frac{eta}{eta_c}$$ and use
          $$frac{dC(eta)}{deta} = frac{d C(eta)}{dx}frac{dx}{deta}=frac{1}{eta_c}frac{d C(eta)}{dx}$$ For sure, your result is correct but, since $k$ has only a few values, you could simplity the final result to get
          $$C'(eta)=frac{1}{eta_c}frac{1+B_1+2 B_2 x+(3 B_3-B_2) x^2-2 B_3 x^3 }{3(1-x)^2}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 8:01









          Claude LeiboviciClaude Leibovici

          126k1158134




          126k1158134























              0












              $begingroup$

              Here's how you could differentiate this particular function:



              $$
              C'(eta) = left[frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}right]'=
              left[frac{1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}right]'=\
              left[left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left[3left(1-frac{eta}{eta_c}right)right]^{-1}right]'=\
              left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)'left[3left(1-frac{eta}{eta_c}right)right]^{-1}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left(left[3left(1-frac{eta}{eta_c}right)right]^{-1}right)'=\
              frac{B_{1}frac{1}{eta_c}+B_{2}frac{2eta}{eta_c^2}+B_{3}frac{3eta^2}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)(-1)left[3left(1-frac{eta}{eta_c}right)right]^{-2}left(3-3frac{eta}{eta_c}right)'=\
              frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{left[3left(1-frac{eta}{eta_c}right)right]^2}cdotfrac{3}{eta_c}=\
              frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3eta_cleft(1-frac{eta}{eta_c}right)^2}.
              $$



              The answer I get appears to be equivalent to what you got.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Here's how you could differentiate this particular function:



                $$
                C'(eta) = left[frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}right]'=
                left[frac{1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}right]'=\
                left[left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left[3left(1-frac{eta}{eta_c}right)right]^{-1}right]'=\
                left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)'left[3left(1-frac{eta}{eta_c}right)right]^{-1}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left(left[3left(1-frac{eta}{eta_c}right)right]^{-1}right)'=\
                frac{B_{1}frac{1}{eta_c}+B_{2}frac{2eta}{eta_c^2}+B_{3}frac{3eta^2}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)(-1)left[3left(1-frac{eta}{eta_c}right)right]^{-2}left(3-3frac{eta}{eta_c}right)'=\
                frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{left[3left(1-frac{eta}{eta_c}right)right]^2}cdotfrac{3}{eta_c}=\
                frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3eta_cleft(1-frac{eta}{eta_c}right)^2}.
                $$



                The answer I get appears to be equivalent to what you got.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Here's how you could differentiate this particular function:



                  $$
                  C'(eta) = left[frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}right]'=
                  left[frac{1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}right]'=\
                  left[left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left[3left(1-frac{eta}{eta_c}right)right]^{-1}right]'=\
                  left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)'left[3left(1-frac{eta}{eta_c}right)right]^{-1}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left(left[3left(1-frac{eta}{eta_c}right)right]^{-1}right)'=\
                  frac{B_{1}frac{1}{eta_c}+B_{2}frac{2eta}{eta_c^2}+B_{3}frac{3eta^2}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)(-1)left[3left(1-frac{eta}{eta_c}right)right]^{-2}left(3-3frac{eta}{eta_c}right)'=\
                  frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{left[3left(1-frac{eta}{eta_c}right)right]^2}cdotfrac{3}{eta_c}=\
                  frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3eta_cleft(1-frac{eta}{eta_c}right)^2}.
                  $$



                  The answer I get appears to be equivalent to what you got.






                  share|cite|improve this answer









                  $endgroup$



                  Here's how you could differentiate this particular function:



                  $$
                  C'(eta) = left[frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}right]'=
                  left[frac{1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}right]'=\
                  left[left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left[3left(1-frac{eta}{eta_c}right)right]^{-1}right]'=\
                  left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)'left[3left(1-frac{eta}{eta_c}right)right]^{-1}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left(left[3left(1-frac{eta}{eta_c}right)right]^{-1}right)'=\
                  frac{B_{1}frac{1}{eta_c}+B_{2}frac{2eta}{eta_c^2}+B_{3}frac{3eta^2}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)(-1)left[3left(1-frac{eta}{eta_c}right)right]^{-2}left(3-3frac{eta}{eta_c}right)'=\
                  frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{left[3left(1-frac{eta}{eta_c}right)right]^2}cdotfrac{3}{eta_c}=\
                  frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3eta_cleft(1-frac{eta}{eta_c}right)^2}.
                  $$



                  The answer I get appears to be equivalent to what you got.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 29 at 19:08









                  Michael RybkinMichael Rybkin

                  4,544522




                  4,544522






























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                      Cabo Verde

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                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                      Gyllenstierna

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