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Thread: Partial derivative chain rule

  1. #1
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    Partial derivative chain rule

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

    $\displaystyle f$ $\displaystyle g$ $\displaystyle f_{x}$ $\displaystyle f_{y}$
    (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:

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

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



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

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

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

    $\displaystyle f$ $\displaystyle g$ $\displaystyle f_{x}$ $\displaystyle f_{y}$
    (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:

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

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



    $\displaystyle \frac{\partial g}{\partial u}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial u} = 4*e^u + 8*e^u = 12$
    You are saying that $\displaystyle f_x= 4$ and $\displaystyle f_y= 8$
    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 $\displaystyle x= e^u+ sin(v)$ and $\displaystyle y= e^u+ cos(v)$, when u= v= 0, $\displaystyle x= e^0+ sin(0)= 1$ and $\displaystyle y=e^0+ cos(0)= 2$. you need to use the (1, 2) row to find $\displaystyle f_x$ and $\displaystyle f_y$.

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

    But the answer in BOB is 7, 2... Where is my error in understanding?
    Last edited by HallsofIvy; Sep 29th 2012 at 08:56 AM.
    Thanks from sgcb
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