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Chain rule (university level, help desperately needed) Watch

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    Here is my knowledge of the chain rule from first year (and I understand it all):

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    In my maths course, I was taught:

    If g: R^n ---> R^m and f: R^m ---> R

    And we have x = g(u) and y = f(x)

    Then dy/dx = (\frac{\partial y}{\partial x_1}, \frac{\partial y}{\partial x_2}, ..., \frac{\partial y}{\partial x_m})

    And dx/du = \displaystyle \begin{pmatrix} \frac{\partial x_1}{\partial u_1} & \frac{\partial x_1}{\partial u_2} & ... & \frac{\partial x_1}{\partial u_n} \\ \frac{\partial x_2}{\partial u_1} & \frac{\partial x_2}{\partial u_2} & ... & \frac{\partial x_2}{\partial u_n} \\ . & . & & . \\ . & . & & .\\ . & . & & . \\ \frac{\partial x_m}{\partial u_1} & \frac{\partial x_m}{\partial u_2} & ... & \frac{\partial x_m}{\partial u_n}\end{pmatrix}

    Therefore, dy/du = dy/dx * dx/du

    Which gives the set of equations

     \frac{\partial y}{\partial u_1} = \frac{\partial y}{\partial x_1}\frac{\partial x_1}{\partial u_1} + \frac{\partial y}{\partial x_2}\frac{\partial x_2}{\partial u_1} + ... + \frac{\partial y}{\partial x_m}\frac{\partial x_m}{\partial u_1}

     \frac{\partial y}{\partial u_2} = \frac{\partial y}{\partial x_1}\frac{\partial x_1}{\partial u_2} + \frac{\partial y}{\partial x_2}\frac{\partial x_2}{\partial u_1} + ... + \frac{\partial y}{\partial x_m}\frac{\partial x_m}{\partial u_2}

    ...

     \frac{\partial y}{\partial u_m} = \frac{\partial y}{\partial x_1}\frac{\partial x_1}{\partial u_m} + \frac{\partial y}{\partial x_2}\frac{\partial x_2}{\partial u_m} + ... + \frac{\partial y}{\partial x_m}\frac{\partial x_m}{\partial u_m}

    This all makes sense to me


    The problem is now applying it to an economics problem. We have that

    p = \begin{pmatrix} p_i \\ p_j \\ p_k \end{pmatrix}

    Then we have three functions:

    H(p, u)
    D(p, y)
    C(p, u)

    u is a constant (but don't throw it out in your working please) and y = C(p, u)

    You are given that H(\mathbf{p}, u) = D(\mathbf{p}, y). Then you replace y with C to get

    H(\mathbf{p}, u) = D(\mathbf{p}, y) = D(\mathbf{p}, C(\mathbf{p}, u))

    i.e. H(\mathbf{p}, u) = D(\mathbf{p}, C(\mathbf{p}, u))

    Now you're asked to differentiate through with respect to p_j

    The answer is H_j(\mathbf{p}, u) = D_j(\mathbf{p}, y) + D_y(\mathbf{p}, y)C_j(\mathbf{p}, u)

    Where H_j means \frac{\partial H}{\partial p_j} and likewise for D_j and C_j.

    Can someone work me through that differentiation in as many small steps as possible? It would really help me out.
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    (Original post by danny111)
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    (Original post by Fellas)
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    Not sure if anyone else will bother hacking through the notation, so hoping you guys can help! I'm trying to prove the Slutsky equation, but I just don't get the chain rule differentiation that we keep doing in this course!
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    (Original post by Swayum)
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    Draw the tree thingy.

    D is a function of p and y

    so you have 2 branches of D, p and y.

    y = C(p,u) so draw two more branches of y for p and u

    so if you differentiate wrt to p_j you get two terms, one straight from the p, and one through y and for this term it's where you use the chain rule. Don't know if that helps with understanding how it relates to the mathematical definition, but I remember this method from MA100 and just used it.
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    Slutsky
    lol
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    (Original post by danny111)
    Draw the tree thingy.

    D is a function of p and y

    so you have 2 branches of D, p and y.

    y = C(p,u) so draw two more branches of y for p and u

    so if you differentiate wrt to p_j you get two terms, one straight from the p, and one through y and for this term it's where you use the chain rule. Don't know if that helps with understanding how it relates to the mathematical definition, but I remember this method from MA100 and just used it.
    I've never seen this tree method. Can you see it anywhere in the MA100 slides or in the calculus book? Would help me a lot if you could reference it for me so I can understand it properly.
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    (Original post by Swayum)
    I've never seen this tree method. Can you see it anywhere in the MA100 slides or in the calculus book? Would help me a lot if you could reference it for me so I can understand it properly.
    .http://people.usd.edu/~jflores/Multi...ain%20Rule.htm
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    (Original post by Swayum)
    I've never seen this tree method. Can you see it anywhere in the MA100 slides or in the calculus book? Would help me a lot if you could reference it for me so I can understand it properly.
    i dont know, i found it somewhere in my notes.

    but rbn linked a good site, so thank you to him.
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    Ok so I make the tree out to be like so:



    So when trying to work out \frac{\partial D}{\partial p_j}

    From the left branch I get dD/dp * dp/dpj = dD/dpj

    From the right branch I get dD/dy * dy/dp * dp/dpj = dD/dy * dy/dpj = D_yC_j

    Add them and you get the right answer.

    Is this the correct way of doing it or I have misunderstood?
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    (Original post by Swayum)
    Ok so I make the tree out to be like so:


    So when trying to work out \frac{\partial D}{\partial p_j}

    From the left branch I get dD/dp * dp/dpj = dD/dpj

    From the right branch I get dD/dy * dy/dp * dp/dpj = dD/dy * dy/dpj = D_yC_j

    Add them and you get the right answer.

    Is this the correct way of doing it or I have misunderstood?
    That's how I do it, and it also works for the supply function of the firm.
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    (Original post by Swayum)

    From the left branch I get dD/dp * dp/dpj = dD/dpj

    From the right branch I get dD/dy * dy/dp * dp/dpj = dD/dy * dy/dpj = D_yC_j

    Add them and you get the right answer.

    Is this the correct way of doing it or I have misunderstood?
    But yes. The slutsky equation is formed of the substitution and income effects. By doing the tree thing it looks like you are working out the separate effects (P, Y) and combining them.

    I have something cool. Just working out how to get it on here:

    http://img220.imageshack.us/i/202py.jpg/

    Also just to add courtesy of Konrad Burchardi.
 
 
 
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