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Math Help - Partial Derivatives Of Composite Function

  1. #1
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    Partial Derivatives Of Composite Function

    Question: Let  x(t_1, t_2) = t_1 e^{t_2}, y(t_1, t_2) = t_1^2 + sin(t_1 t_2) . Let  f(x,y) be a differentiable function  f: \mathbb{R}^2 \rightarrow \mathbb{R} . Let  g(t_1, t_2) = f( x(t_1, t_2), y(t_1, t_2) ) . Express  \frac{ \partial g}{\partial t_1} (1,0) in terms of partial derivatives of  f .

    My Solution: Let  h: \mathbb{R}^2 \rightarrow \mathbb{R}^2 be defined as  h(t_1, t_2) = (x(t_1,t_2), y(t_1, t_2)) = (t_1e^{t_2}, t_1^2 + sin(t_1 t_2)) . Then  g(t_1, t_2) = f(x(t_1, t_2), y(t_1, t_2)) = f(h(t_1, t_2)) . Note that  f is differentiable and so is  h , because its component functions are differentiable. Thus their composition,  f(h(t_1,t_2)) = g(t_1, t_2) is differentiable. Note that  dfh(1,0) = df(h(1,0)) dh(1,0) . Observe that  dfh(1,0) = \begin{pmatrix} \frac{\partial fh}{\partial t_1}(1, 0) & \frac{\partial fh}{\partial t_2} (1,0) \end{pmatrix} .


    We also know that  df(h(1,0)) = \begin{pmatrix} \frac{\partial f}{\partial x} h(1,0) & \frac{\partial f}{\partial y} h(1,0) \end{pmatrix} and dh(1,0) = \begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix} .

    Thus  \begin{pmatrix} \frac{\partial fh}{\partial t_1}(1,0) & \frac{\partial fh}{\partial t_2} (1,0) \end{pmatrix} = \begin{pmatrix} \frac{\partial f}{\partial x} h(1,0) & \frac{\partial f}{\partial y} h(1,0) \end{pmatrix} \begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}

     = \begin{pmatrix}  \frac{\partial f}{\partial x}(1,0) + \frac{\partial  f}{\partial y}(1,0) & 2 \frac{\partial f}{\partial x}(1,0) +  \frac{\partial f}{\partial y}(1,0) \end{pmatrix}

    Thus  \frac{\partial g}{\partial t_1}(1,0) = \frac{\partial fh}{\partial t_1}(1,0) = \frac{\partial f}{\partial x} (1,0) + \frac{\partial f}{\partial y} (1,0) .

    Is this correct? Also, is there a faster way to do this? Also, one last question...are the partial derivatives of f taken with respect to x and y or with respect to t_1 and t_2 ? Or are they just dummy variables that I can interchange?
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  2. #2
    Member HappyJoe's Avatar
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    It looks almost perfect. It sounds like you know what's going on. I didn't check you calculation of dh(1,0), but this is not the hardest part anyway.

    Quote Originally Posted by JG89 View Post
    Thus  \begin{pmatrix} \frac{\partial fh}{\partial t_1}(1,0) & \frac{\partial fh}{\partial t_2} (1,0) \end{pmatrix} = \begin{pmatrix} \frac{\partial f}{\partial x} h(1,0) & \frac{\partial f}{\partial y} h(1,0) \end{pmatrix} \begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}

     = \begin{pmatrix}  \frac{\partial f}{\partial x}(1,0) + \frac{\partial  f}{\partial y}(1,0) & 2 \frac{\partial f}{\partial x}(1,0) +  \frac{\partial f}{\partial y}(1,0) \end{pmatrix}

    Thus  \frac{\partial g}{\partial t_1}(1,0) = \frac{\partial fh}{\partial t_1}(1,0) = \frac{\partial f}{\partial x} (1,0) + \frac{\partial f}{\partial y} (1,0) .

    Is this correct? Also, is there a faster way to do this? Also, one last question...are the partial derivatives of f taken with respect to x and y or with respect to t_1 and t_2 ? Or are they just dummy variables that I can interchange?
    Note that in the expression

     \begin{pmatrix}  \frac{\partial f}{\partial x}(1,0) +  \frac{\partial  f}{\partial y}(1,0) & 2 \frac{\partial f}{\partial  x}(1,0) +  \frac{\partial f}{\partial y}(1,0) \end{pmatrix}

    you forget to apply h to (1,0) (so that instead of e.g.  \frac{\partial f}{\partial x}(1,0), you want \frac{\partial f}{\partial x}h(1,0) ).

    As for a faster way of solving this, I guess you could save a wee bit of time, but it's not a method I would utilize in a small problem like this. Specifically, you are being asked only for \frac{\partial g}{\partial t_1}. For calculating this, you do not need the second column of the matrix for dh(1,0), so you could save yourself some work by calculating only the first column.

    In this problem, the partial derivatives of f are being taken with respect to the variables x and y. Since x and y are, in turn, functions of the variables t_1,t_2, you could also consider f as a function of t_1 and t_2. But the partial derivatives of f with respect to t_1 and t_2 will be completely different from the partial derivatives of f with respect to x and y.
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