Partial derivative with constant












0












$begingroup$


Given



$$c = 0.03 + 0.08a$$



What is $dfrac{partial c}{partial a}$?



I'm guessing $dfrac{partial c}{partial a} = 0.03 + 0.08 = 0.11$?
The constant is confusing me. Thanks.






partial-derivative







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  • 2




    $begingroup$
    The partial derivative of a constant is the same as the ordinary derivative of a constant, namely...?
    $endgroup$
    – David
    May 30 '14 at 5:38






  • 3




    $begingroup$
    0? So dc/da = 0.08? That's what you are saying?
    $endgroup$
    – user122415
    May 30 '14 at 5:40






  • 2




    $begingroup$
    @user122415 Bingo! You may now post an answer to your own question!
    $endgroup$
    – 200_success
    May 30 '14 at 5:44
















0












$begingroup$


Given



$$c = 0.03 + 0.08a$$



What is $dfrac{partial c}{partial a}$?



I'm guessing $dfrac{partial c}{partial a} = 0.03 + 0.08 = 0.11$?
The constant is confusing me. Thanks.






partial-derivative







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The partial derivative of a constant is the same as the ordinary derivative of a constant, namely...?
    $endgroup$
    – David
    May 30 '14 at 5:38






  • 3




    $begingroup$
    0? So dc/da = 0.08? That's what you are saying?
    $endgroup$
    – user122415
    May 30 '14 at 5:40






  • 2




    $begingroup$
    @user122415 Bingo! You may now post an answer to your own question!
    $endgroup$
    – 200_success
    May 30 '14 at 5:44














0












0








0





$begingroup$


Given



$$c = 0.03 + 0.08a$$



What is $dfrac{partial c}{partial a}$?



I'm guessing $dfrac{partial c}{partial a} = 0.03 + 0.08 = 0.11$?
The constant is confusing me. Thanks.






partial-derivative







share|cite|improve this question











$endgroup$




Given



$$c = 0.03 + 0.08a$$



What is $dfrac{partial c}{partial a}$?



I'm guessing $dfrac{partial c}{partial a} = 0.03 + 0.08 = 0.11$?
The constant is confusing me. Thanks.






partial-derivative




partial-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 30 '14 at 5:44









200_success

668515




668515










asked May 30 '14 at 5:36









user122415user122415

761314




761314








  • 2




    $begingroup$
    The partial derivative of a constant is the same as the ordinary derivative of a constant, namely...?
    $endgroup$
    – David
    May 30 '14 at 5:38






  • 3




    $begingroup$
    0? So dc/da = 0.08? That's what you are saying?
    $endgroup$
    – user122415
    May 30 '14 at 5:40






  • 2




    $begingroup$
    @user122415 Bingo! You may now post an answer to your own question!
    $endgroup$
    – 200_success
    May 30 '14 at 5:44














  • 2




    $begingroup$
    The partial derivative of a constant is the same as the ordinary derivative of a constant, namely...?
    $endgroup$
    – David
    May 30 '14 at 5:38






  • 3




    $begingroup$
    0? So dc/da = 0.08? That's what you are saying?
    $endgroup$
    – user122415
    May 30 '14 at 5:40






  • 2




    $begingroup$
    @user122415 Bingo! You may now post an answer to your own question!
    $endgroup$
    – 200_success
    May 30 '14 at 5:44








2




2




$begingroup$
The partial derivative of a constant is the same as the ordinary derivative of a constant, namely...?
$endgroup$
– David
May 30 '14 at 5:38




$begingroup$
The partial derivative of a constant is the same as the ordinary derivative of a constant, namely...?
$endgroup$
– David
May 30 '14 at 5:38




3




3




$begingroup$
0? So dc/da = 0.08? That's what you are saying?
$endgroup$
– user122415
May 30 '14 at 5:40




$begingroup$
0? So dc/da = 0.08? That's what you are saying?
$endgroup$
– user122415
May 30 '14 at 5:40




2




2




$begingroup$
@user122415 Bingo! You may now post an answer to your own question!
$endgroup$
– 200_success
May 30 '14 at 5:44




$begingroup$
@user122415 Bingo! You may now post an answer to your own question!
$endgroup$
– 200_success
May 30 '14 at 5:44










1 Answer
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$begingroup$

Think of $c$ as $c(a)$, as in $c$ is a function in terms of the independent variable $a$. (This means $c$ is not a constant.) Hopefully this will help clear that confusion you had with the $c$.



Given $c = 0.03 + 0.08a$



Then begin{align}frac{partial c}{partial a} &= frac{partial}{partial a} (0.03 + 0.08a) \ &=frac{partial}{partial a}(0.03) + frac{partial}{partial a}(0.08a) \ &= 0 + 0.08 \ &=boxed{0.08}end{align}






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    1 Answer
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    1 Answer
    1






    active

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    0












    $begingroup$

    Think of $c$ as $c(a)$, as in $c$ is a function in terms of the independent variable $a$. (This means $c$ is not a constant.) Hopefully this will help clear that confusion you had with the $c$.



    Given $c = 0.03 + 0.08a$



    Then begin{align}frac{partial c}{partial a} &= frac{partial}{partial a} (0.03 + 0.08a) \ &=frac{partial}{partial a}(0.03) + frac{partial}{partial a}(0.08a) \ &= 0 + 0.08 \ &=boxed{0.08}end{align}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Think of $c$ as $c(a)$, as in $c$ is a function in terms of the independent variable $a$. (This means $c$ is not a constant.) Hopefully this will help clear that confusion you had with the $c$.



      Given $c = 0.03 + 0.08a$



      Then begin{align}frac{partial c}{partial a} &= frac{partial}{partial a} (0.03 + 0.08a) \ &=frac{partial}{partial a}(0.03) + frac{partial}{partial a}(0.08a) \ &= 0 + 0.08 \ &=boxed{0.08}end{align}






      share|cite|improve this answer









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        0








        0





        $begingroup$

        Think of $c$ as $c(a)$, as in $c$ is a function in terms of the independent variable $a$. (This means $c$ is not a constant.) Hopefully this will help clear that confusion you had with the $c$.



        Given $c = 0.03 + 0.08a$



        Then begin{align}frac{partial c}{partial a} &= frac{partial}{partial a} (0.03 + 0.08a) \ &=frac{partial}{partial a}(0.03) + frac{partial}{partial a}(0.08a) \ &= 0 + 0.08 \ &=boxed{0.08}end{align}






        share|cite|improve this answer









        $endgroup$



        Think of $c$ as $c(a)$, as in $c$ is a function in terms of the independent variable $a$. (This means $c$ is not a constant.) Hopefully this will help clear that confusion you had with the $c$.



        Given $c = 0.03 + 0.08a$



        Then begin{align}frac{partial c}{partial a} &= frac{partial}{partial a} (0.03 + 0.08a) \ &=frac{partial}{partial a}(0.03) + frac{partial}{partial a}(0.08a) \ &= 0 + 0.08 \ &=boxed{0.08}end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 30 '14 at 6:11









        CookieCookie

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