Results 1 to 7 of 7

Math Help - partial derivative without knowing actual function?

  1. #1
    Member garymarkhov's Avatar
    Joined
    Aug 2009
    Posts
    149
    Awards
    1

    partial derivative without knowing actual function?

    My textbook, Advanced Macroeconomics (Romer), introduces a function Y=F(K,AL), which can be written \frac{Y}{AL}=F(\frac{K}{AL},1) and is often shortened to y=f(k). It then suggests that \frac{\partial F(K,AL)}{\partial K} = ALf'(\frac{K}{AL})(\frac{1}{AL}).

    But how can they take the partial with respect to K if they don't specify the actual function? They say what the function depends on, but not what the actual function is. What is happening here?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    Looks like they used the Chain Rule:

    \frac{\partial}{\partial K}\left(AL\cdot F\left(\frac{K}{AL},1\right)\right)=AL\frac{\parti  al F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{\partial}{  \partial K}\left(\frac{K}{AL}\right)=AL\frac{\partial F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{1}{AL}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member garymarkhov's Avatar
    Joined
    Aug 2009
    Posts
    149
    Awards
    1
    Quote Originally Posted by Scott H View Post
    Looks like they used the Chain Rule:

    \frac{\partial}{\partial K}\left(AL\cdot F\left(\frac{K}{AL},1\right)\right)=AL\frac{\parti  al F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{\partial}{  \partial K}\left(\frac{K}{AL}\right)=AL\frac{\partial F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{1}{AL}.
    That looks about right. A little strange to find the partial without knowing what the actual function is, but I guess it would work out if we plugged in the actual function?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    Correct, as the Chain Rule applies to every function.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member garymarkhov's Avatar
    Joined
    Aug 2009
    Posts
    149
    Awards
    1
    Quote Originally Posted by Scott H View Post
    Correct, as the Chain Rule applies to every function.
    Hmm, so say you discovered that the function is Y=F(K,AL)=K^\alpha(AL)^{1-\alpha}. Can we figure out what the partial with respect to K is from f'(\frac{K}{AL})?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    Yes.

    Differentiating the function normally, we obtain

    \frac{\partial}{\partial K}(K^{\alpha}(AL)^{1-\alpha})=\alpha K^{\alpha-1}(AL)^{1-\alpha}.

    Using the formula derived from the Chain Rule, we obtain

    \begin{aligned}<br />
\frac{\partial}{\partial K}\left(AL\cdot F\left(\frac{K}{AL},1\right)\right)&=AL\frac{\part  ial F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{1}{AL}\\<br />
&=AL\alpha\left(\frac{K}{AL}\right)^{\alpha-1}(1)^{1-\alpha}\cdot\frac{1}{AL}\\<br />
&=\frac{\alpha K^{\alpha-1}}{(AL)^{\alpha-1}}\\<br />
&=\alpha K^{\alpha-1}(AL)^{1-\alpha}.<br />
\end{aligned}<br />
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member garymarkhov's Avatar
    Joined
    Aug 2009
    Posts
    149
    Awards
    1
    That's great, thanks for your help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 17th 2011, 05:47 AM
  2. Replies: 1
    Last Post: April 23rd 2011, 11:32 PM
  3. Replies: 1
    Last Post: May 4th 2010, 01:49 PM
  4. Partial derivative of a function
    Posted in the Calculus Forum
    Replies: 8
    Last Post: March 13th 2010, 05:09 AM
  5. Replies: 3
    Last Post: December 4th 2008, 01:09 AM

Search Tags


/mathhelpforum @mathhelpforum