Find $frac{partial}{partial p}$ and $frac{partial}{partial q}$ of $U(f(p) + q^2, g(q)^2)$












1












$begingroup$


Here's my question:



Suppose $U(x_1, x_2)$ is a differentiable function. Suppose $f(p)$ and $g(q)$ are differentiable functions that depend only on $p$ and $q$ respectively.



Find $frac{partial}{partial p}$ and $frac{partial}{partial q}$ of $U(f(p) + q^2, g(q)^2).$



Be sure to include the value at which each partial derivative is evaluated.



I know the chain rule for multivariable functions has the general look of:



$frac{partial f}{partial t_j} = frac{partial f}{partial x_1}frac{partial x_1}{partial t_j} + frac{partial f}{partial x_2}frac{partial x_2}{partial t_j} + ... + frac{partial f}{partial x_n}frac{partial x_n}{partial t_j}$



I understand that $p$ and $q$ are supposed to be my $t_1$ and $t_2$, and that $x_1$ and $x_2$ are the same. I'm not sure how to apply the chain rule to this though, since $x_1$ is a function of two variables but $x_2$ is a function of just one.



Any ideas of what I'm supposed to do?



Thanks ahead of time.






multivariable-calculus optimization economics chain-rule







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Here's my question:



    Suppose $U(x_1, x_2)$ is a differentiable function. Suppose $f(p)$ and $g(q)$ are differentiable functions that depend only on $p$ and $q$ respectively.



    Find $frac{partial}{partial p}$ and $frac{partial}{partial q}$ of $U(f(p) + q^2, g(q)^2).$



    Be sure to include the value at which each partial derivative is evaluated.



    I know the chain rule for multivariable functions has the general look of:



    $frac{partial f}{partial t_j} = frac{partial f}{partial x_1}frac{partial x_1}{partial t_j} + frac{partial f}{partial x_2}frac{partial x_2}{partial t_j} + ... + frac{partial f}{partial x_n}frac{partial x_n}{partial t_j}$



    I understand that $p$ and $q$ are supposed to be my $t_1$ and $t_2$, and that $x_1$ and $x_2$ are the same. I'm not sure how to apply the chain rule to this though, since $x_1$ is a function of two variables but $x_2$ is a function of just one.



    Any ideas of what I'm supposed to do?



    Thanks ahead of time.






    multivariable-calculus optimization economics chain-rule







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Here's my question:



      Suppose $U(x_1, x_2)$ is a differentiable function. Suppose $f(p)$ and $g(q)$ are differentiable functions that depend only on $p$ and $q$ respectively.



      Find $frac{partial}{partial p}$ and $frac{partial}{partial q}$ of $U(f(p) + q^2, g(q)^2).$



      Be sure to include the value at which each partial derivative is evaluated.



      I know the chain rule for multivariable functions has the general look of:



      $frac{partial f}{partial t_j} = frac{partial f}{partial x_1}frac{partial x_1}{partial t_j} + frac{partial f}{partial x_2}frac{partial x_2}{partial t_j} + ... + frac{partial f}{partial x_n}frac{partial x_n}{partial t_j}$



      I understand that $p$ and $q$ are supposed to be my $t_1$ and $t_2$, and that $x_1$ and $x_2$ are the same. I'm not sure how to apply the chain rule to this though, since $x_1$ is a function of two variables but $x_2$ is a function of just one.



      Any ideas of what I'm supposed to do?



      Thanks ahead of time.






      multivariable-calculus optimization economics chain-rule







      share|cite|improve this question









      $endgroup$




      Here's my question:



      Suppose $U(x_1, x_2)$ is a differentiable function. Suppose $f(p)$ and $g(q)$ are differentiable functions that depend only on $p$ and $q$ respectively.



      Find $frac{partial}{partial p}$ and $frac{partial}{partial q}$ of $U(f(p) + q^2, g(q)^2).$



      Be sure to include the value at which each partial derivative is evaluated.



      I know the chain rule for multivariable functions has the general look of:



      $frac{partial f}{partial t_j} = frac{partial f}{partial x_1}frac{partial x_1}{partial t_j} + frac{partial f}{partial x_2}frac{partial x_2}{partial t_j} + ... + frac{partial f}{partial x_n}frac{partial x_n}{partial t_j}$



      I understand that $p$ and $q$ are supposed to be my $t_1$ and $t_2$, and that $x_1$ and $x_2$ are the same. I'm not sure how to apply the chain rule to this though, since $x_1$ is a function of two variables but $x_2$ is a function of just one.



      Any ideas of what I'm supposed to do?



      Thanks ahead of time.






      multivariable-calculus optimization economics chain-rule




      multivariable-calculus optimization economics chain-rule






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 27 at 20:33









      user591271user591271

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          $begingroup$

          First note that you can defined the functions : $x_1 : (p,q) mapsto f(p)+q^2$ and $x_2 :(p,q) mapsto g(q)^2$. Then we can defined : $G : (p,q) mapsto U(x_1(p,q), x_2(p,q))$.



          So we want to calculate : $frac{partial G}{partial p}$. To do so let's use the chain rule :



          $$frac{partial G}{partial p}(p,q) = f'(p) frac{partial U}{partial x}(x_1(p,q), x_2(p,q))$$



          Note that since $f$ is a function of one variable we have : $f'(p) = frac{partial f}{partial x}(p)$..



          We also get using once again the chain rule :



          $$frac{partial G}{partial q} = 2q frac{partial U}{partial x} + 2g'(q)g(q) frac{partial U}{partial y}$$






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            $begingroup$

            First note that you can defined the functions : $x_1 : (p,q) mapsto f(p)+q^2$ and $x_2 :(p,q) mapsto g(q)^2$. Then we can defined : $G : (p,q) mapsto U(x_1(p,q), x_2(p,q))$.



            So we want to calculate : $frac{partial G}{partial p}$. To do so let's use the chain rule :



            $$frac{partial G}{partial p}(p,q) = f'(p) frac{partial U}{partial x}(x_1(p,q), x_2(p,q))$$



            Note that since $f$ is a function of one variable we have : $f'(p) = frac{partial f}{partial x}(p)$..



            We also get using once again the chain rule :



            $$frac{partial G}{partial q} = 2q frac{partial U}{partial x} + 2g'(q)g(q) frac{partial U}{partial y}$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              First note that you can defined the functions : $x_1 : (p,q) mapsto f(p)+q^2$ and $x_2 :(p,q) mapsto g(q)^2$. Then we can defined : $G : (p,q) mapsto U(x_1(p,q), x_2(p,q))$.



              So we want to calculate : $frac{partial G}{partial p}$. To do so let's use the chain rule :



              $$frac{partial G}{partial p}(p,q) = f'(p) frac{partial U}{partial x}(x_1(p,q), x_2(p,q))$$



              Note that since $f$ is a function of one variable we have : $f'(p) = frac{partial f}{partial x}(p)$..



              We also get using once again the chain rule :



              $$frac{partial G}{partial q} = 2q frac{partial U}{partial x} + 2g'(q)g(q) frac{partial U}{partial y}$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                First note that you can defined the functions : $x_1 : (p,q) mapsto f(p)+q^2$ and $x_2 :(p,q) mapsto g(q)^2$. Then we can defined : $G : (p,q) mapsto U(x_1(p,q), x_2(p,q))$.



                So we want to calculate : $frac{partial G}{partial p}$. To do so let's use the chain rule :



                $$frac{partial G}{partial p}(p,q) = f'(p) frac{partial U}{partial x}(x_1(p,q), x_2(p,q))$$



                Note that since $f$ is a function of one variable we have : $f'(p) = frac{partial f}{partial x}(p)$..



                We also get using once again the chain rule :



                $$frac{partial G}{partial q} = 2q frac{partial U}{partial x} + 2g'(q)g(q) frac{partial U}{partial y}$$






                share|cite|improve this answer









                $endgroup$



                First note that you can defined the functions : $x_1 : (p,q) mapsto f(p)+q^2$ and $x_2 :(p,q) mapsto g(q)^2$. Then we can defined : $G : (p,q) mapsto U(x_1(p,q), x_2(p,q))$.



                So we want to calculate : $frac{partial G}{partial p}$. To do so let's use the chain rule :



                $$frac{partial G}{partial p}(p,q) = f'(p) frac{partial U}{partial x}(x_1(p,q), x_2(p,q))$$



                Note that since $f$ is a function of one variable we have : $f'(p) = frac{partial f}{partial x}(p)$..



                We also get using once again the chain rule :



                $$frac{partial G}{partial q} = 2q frac{partial U}{partial x} + 2g'(q)g(q) frac{partial U}{partial y}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 27 at 22:39









                ThinkingThinking

                1,26816




                1,26816






























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