Partial derivative 8












2












$begingroup$


I have the following derivative:
$$f'(k)=frac{a(1+bk)}{k^{1-a}(1+abk)^{^{a}}}$$
and have to compute the partial derivative with respect to b, possibly obtaining the following result:



$$frac{partial f'(k)}{partial b}=frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}$$



I've tried to do the exercise but I cannot get closer to the desired result than this:



$$frac{partial f'{(k)}}{partial b}=ak^{a}(1+abk)^{-a}left [ 1-frac{a(a+abk)}{(1+abk)} right ]$$



Thank you so much and Happy Christmas! :)






calculus derivatives partial-derivative







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  • 1




    $begingroup$
    Use logarithmic differentiation just as I made it in your previous post and factor terms..
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 14:54












  • $begingroup$
    Thank u so much! I will try that way. What is the advantage of log differentiation? It is possible to obtain the same result with normal derivation?
    $endgroup$
    – Alessandro
    Dec 25 '18 at 14:57






  • 1




    $begingroup$
    The advantage is that evrything becomes simpler. For sure, using normal derivation will give the same answer (but more tedious).
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 14:58










  • $begingroup$
    In this case since k is now a constant in the first step i have to start with this, right? $$frac{y'}{y}=frac{k}{(1+bk)}-frac{a^{2}k}{(1+abk)}$$
    $endgroup$
    – Alessandro
    Dec 25 '18 at 15:30


















2












$begingroup$


I have the following derivative:
$$f'(k)=frac{a(1+bk)}{k^{1-a}(1+abk)^{^{a}}}$$
and have to compute the partial derivative with respect to b, possibly obtaining the following result:



$$frac{partial f'(k)}{partial b}=frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}$$



I've tried to do the exercise but I cannot get closer to the desired result than this:



$$frac{partial f'{(k)}}{partial b}=ak^{a}(1+abk)^{-a}left [ 1-frac{a(a+abk)}{(1+abk)} right ]$$



Thank you so much and Happy Christmas! :)






calculus derivatives partial-derivative







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Use logarithmic differentiation just as I made it in your previous post and factor terms..
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 14:54












  • $begingroup$
    Thank u so much! I will try that way. What is the advantage of log differentiation? It is possible to obtain the same result with normal derivation?
    $endgroup$
    – Alessandro
    Dec 25 '18 at 14:57






  • 1




    $begingroup$
    The advantage is that evrything becomes simpler. For sure, using normal derivation will give the same answer (but more tedious).
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 14:58










  • $begingroup$
    In this case since k is now a constant in the first step i have to start with this, right? $$frac{y'}{y}=frac{k}{(1+bk)}-frac{a^{2}k}{(1+abk)}$$
    $endgroup$
    – Alessandro
    Dec 25 '18 at 15:30
















2












2








2





$begingroup$


I have the following derivative:
$$f'(k)=frac{a(1+bk)}{k^{1-a}(1+abk)^{^{a}}}$$
and have to compute the partial derivative with respect to b, possibly obtaining the following result:



$$frac{partial f'(k)}{partial b}=frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}$$



I've tried to do the exercise but I cannot get closer to the desired result than this:



$$frac{partial f'{(k)}}{partial b}=ak^{a}(1+abk)^{-a}left [ 1-frac{a(a+abk)}{(1+abk)} right ]$$



Thank you so much and Happy Christmas! :)






calculus derivatives partial-derivative







share|cite|improve this question









$endgroup$




I have the following derivative:
$$f'(k)=frac{a(1+bk)}{k^{1-a}(1+abk)^{^{a}}}$$
and have to compute the partial derivative with respect to b, possibly obtaining the following result:



$$frac{partial f'(k)}{partial b}=frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}$$



I've tried to do the exercise but I cannot get closer to the desired result than this:



$$frac{partial f'{(k)}}{partial b}=ak^{a}(1+abk)^{-a}left [ 1-frac{a(a+abk)}{(1+abk)} right ]$$



Thank you so much and Happy Christmas! :)






calculus derivatives partial-derivative




calculus derivatives partial-derivative






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share|cite|improve this question











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share|cite|improve this question










asked Dec 25 '18 at 14:50









AlessandroAlessandro

277




277








  • 1




    $begingroup$
    Use logarithmic differentiation just as I made it in your previous post and factor terms..
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 14:54












  • $begingroup$
    Thank u so much! I will try that way. What is the advantage of log differentiation? It is possible to obtain the same result with normal derivation?
    $endgroup$
    – Alessandro
    Dec 25 '18 at 14:57






  • 1




    $begingroup$
    The advantage is that evrything becomes simpler. For sure, using normal derivation will give the same answer (but more tedious).
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 14:58










  • $begingroup$
    In this case since k is now a constant in the first step i have to start with this, right? $$frac{y'}{y}=frac{k}{(1+bk)}-frac{a^{2}k}{(1+abk)}$$
    $endgroup$
    – Alessandro
    Dec 25 '18 at 15:30
















  • 1




    $begingroup$
    Use logarithmic differentiation just as I made it in your previous post and factor terms..
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 14:54












  • $begingroup$
    Thank u so much! I will try that way. What is the advantage of log differentiation? It is possible to obtain the same result with normal derivation?
    $endgroup$
    – Alessandro
    Dec 25 '18 at 14:57






  • 1




    $begingroup$
    The advantage is that evrything becomes simpler. For sure, using normal derivation will give the same answer (but more tedious).
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 14:58










  • $begingroup$
    In this case since k is now a constant in the first step i have to start with this, right? $$frac{y'}{y}=frac{k}{(1+bk)}-frac{a^{2}k}{(1+abk)}$$
    $endgroup$
    – Alessandro
    Dec 25 '18 at 15:30










1




1




$begingroup$
Use logarithmic differentiation just as I made it in your previous post and factor terms..
$endgroup$
– Claude Leibovici
Dec 25 '18 at 14:54






$begingroup$
Use logarithmic differentiation just as I made it in your previous post and factor terms..
$endgroup$
– Claude Leibovici
Dec 25 '18 at 14:54














$begingroup$
Thank u so much! I will try that way. What is the advantage of log differentiation? It is possible to obtain the same result with normal derivation?
$endgroup$
– Alessandro
Dec 25 '18 at 14:57




$begingroup$
Thank u so much! I will try that way. What is the advantage of log differentiation? It is possible to obtain the same result with normal derivation?
$endgroup$
– Alessandro
Dec 25 '18 at 14:57




1




1




$begingroup$
The advantage is that evrything becomes simpler. For sure, using normal derivation will give the same answer (but more tedious).
$endgroup$
– Claude Leibovici
Dec 25 '18 at 14:58




$begingroup$
The advantage is that evrything becomes simpler. For sure, using normal derivation will give the same answer (but more tedious).
$endgroup$
– Claude Leibovici
Dec 25 '18 at 14:58












$begingroup$
In this case since k is now a constant in the first step i have to start with this, right? $$frac{y'}{y}=frac{k}{(1+bk)}-frac{a^{2}k}{(1+abk)}$$
$endgroup$
– Alessandro
Dec 25 '18 at 15:30






$begingroup$
In this case since k is now a constant in the first step i have to start with this, right? $$frac{y'}{y}=frac{k}{(1+bk)}-frac{a^{2}k}{(1+abk)}$$
$endgroup$
– Alessandro
Dec 25 '18 at 15:30












1 Answer
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$begingroup$

Your answer is correct. It's just a matter of simplifications.
begin{align}
frac{partial f'(k)}{partial b}
& = frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}\
& = frac{ak^{a}(1+a+abk-a-a^2-a^2bk)}{(1+abk)^{a+1}}\
& = frac{ak^{a}(1+abk-a^2-a^2bk)}{(1+abk)^{a+1}}\
&=frac{ak^{a}(1+abk)^{-a}(1+abk-a^2-a^2bk)}{(1+abk)}\
& = ak^{a}(1+abk)^{-a}left [ 1-frac{a(a+abk)}{(1+abk)} right ]
end{align}






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    $begingroup$

    Your answer is correct. It's just a matter of simplifications.
    begin{align}
    frac{partial f'(k)}{partial b}
    & = frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}\
    & = frac{ak^{a}(1+a+abk-a-a^2-a^2bk)}{(1+abk)^{a+1}}\
    & = frac{ak^{a}(1+abk-a^2-a^2bk)}{(1+abk)^{a+1}}\
    &=frac{ak^{a}(1+abk)^{-a}(1+abk-a^2-a^2bk)}{(1+abk)}\
    & = ak^{a}(1+abk)^{-a}left [ 1-frac{a(a+abk)}{(1+abk)} right ]
    end{align}






    share|cite|improve this answer









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      1












      $begingroup$

      Your answer is correct. It's just a matter of simplifications.
      begin{align}
      frac{partial f'(k)}{partial b}
      & = frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}\
      & = frac{ak^{a}(1+a+abk-a-a^2-a^2bk)}{(1+abk)^{a+1}}\
      & = frac{ak^{a}(1+abk-a^2-a^2bk)}{(1+abk)^{a+1}}\
      &=frac{ak^{a}(1+abk)^{-a}(1+abk-a^2-a^2bk)}{(1+abk)}\
      & = ak^{a}(1+abk)^{-a}left [ 1-frac{a(a+abk)}{(1+abk)} right ]
      end{align}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Your answer is correct. It's just a matter of simplifications.
        begin{align}
        frac{partial f'(k)}{partial b}
        & = frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}\
        & = frac{ak^{a}(1+a+abk-a-a^2-a^2bk)}{(1+abk)^{a+1}}\
        & = frac{ak^{a}(1+abk-a^2-a^2bk)}{(1+abk)^{a+1}}\
        &=frac{ak^{a}(1+abk)^{-a}(1+abk-a^2-a^2bk)}{(1+abk)}\
        & = ak^{a}(1+abk)^{-a}left [ 1-frac{a(a+abk)}{(1+abk)} right ]
        end{align}






        share|cite|improve this answer









        $endgroup$



        Your answer is correct. It's just a matter of simplifications.
        begin{align}
        frac{partial f'(k)}{partial b}
        & = frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}\
        & = frac{ak^{a}(1+a+abk-a-a^2-a^2bk)}{(1+abk)^{a+1}}\
        & = frac{ak^{a}(1+abk-a^2-a^2bk)}{(1+abk)^{a+1}}\
        &=frac{ak^{a}(1+abk)^{-a}(1+abk-a^2-a^2bk)}{(1+abk)}\
        & = ak^{a}(1+abk)^{-a}left [ 1-frac{a(a+abk)}{(1+abk)} right ]
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 15:33









        Thomas ShelbyThomas Shelby

        3,5292525




        3,5292525






























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