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Math Help - Partial derivative chain rule

  1. #1
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    Partial derivative chain rule

    Q: Suppose f is a differentiable function of x and y, and g(u, v) = f(e^u + sin(v), e^u + cos(v)). Use the table of values to calculate g_{u}(0, 0) and g_{v}(0, 0).

    f g f_{x} f_{y}
    (0, 0) 3 6 4 8
    (1, 2) 6 3 2 5

    I apply the chain rule for 2 independent/2 dependent variable how I /think/ it's suppose to be applied in this problem:

    \frac{\partial x}{\partial u}=e^u, \frac{\partial x}{\partial v}=cos(v)

    \frac{\partial y}{\partial u}=e^u, \frac{\partial y}{\partial v}=-sin(v)



    \frac{\partial g}{\partial u}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial u} = 4*e^u + 8*e^u = 12

    \frac{\partial g}{\partial v}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial v} = 4cos(v) - 8sin(v) = 4

    But the answer in BOB is 7, 2... Where is my error in understanding?
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  2. #2
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    Re: Partial derivative chain rule

    Quote Originally Posted by sgcb View Post
    Q: Suppose f is a differentiable function of x and y, and g(u, v) = f(e^u + sin(v), e^u + cos(v)). Use the table of values to calculate g_{u}(0, 0) and g_{v}(0, 0).

    f g f_{x} f_{y}
    (0, 0) 3 6 4 8
    (1, 2) 6 3 2 5

    I apply the chain rule for 2 independent/2 dependent variable how I /think/ it's suppose to be applied in this problem:

    \frac{\partial x}{\partial u}=e^u, \frac{\partial x}{\partial v}=cos(v)

    \frac{\partial y}{\partial u}=e^u, \frac{\partial y}{\partial v}=-sin(v)



    \frac{\partial g}{\partial u}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial u} = 4*e^u + 8*e^u = 12
    You are saying that f_x= 4 and f_y= 8
    Those are the values given in your table (nicely done, by the way) for (0, 0). But you are asked to find the values at u= 0, v= 0. Because x= e^u+ sin(v) and y= e^u+ cos(v), when u= v= 0, x= e^0+ sin(0)= 1 and y=e^0+ cos(0)= 2. you need to use the (1, 2) row to find f_x and f_y.

    \frac{\partial g}{\partial v}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial v} = 4cos(v) - 8sin(v) = 4

    But the answer in BOB is 7, 2... Where is my error in understanding?
    Last edited by HallsofIvy; September 29th 2012 at 09:56 AM.
    Thanks from sgcb
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