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Urgend Calculus Question: Please Look

  1. Sep 24, 2006 #1
    Hi

    Given [tex]z = sin(x + sin(t))[/tex]

    show that [tex]\frac{\partial z}{\partial x} \cdot \frac{\partial ^2 x}{\partial x \partial z} = \frac{\partial z}{\partial t} \cdot \frac{\partial ^2 z} {\partial x^2}[/tex]

    By using the chain-rule I get:

    [tex]f_x(x,t) = cos(x + sin(1))[/tex]

    [tex]f_{xx}(x,t) = -sin(x + sin(1))[/tex]

    [tex]f_t(x,t) = cos(1) \cdot cos(x + sin(1))[/tex]

    [tex]f_{tt}(x,t) = 0[/tex]

    Therefore

    [tex]\frac{\partial z}{\partial x} \cdot \frac{\partial ^2 x}{\partial x \partial z} = cos(x + sin(1)) \cdot 0 = cos(1) \cdot cos(x + sin(1)) \cdot 0 = \frac{\partial z}{\partial t} \cdot \frac{\partial ^2 z} {\partial x^2}[/tex]

    Does that look right ?

    Sincerely
    MM20
     
    Last edited: Sep 24, 2006
  2. jcsd
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