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Math Help - check derivative

  1. #1
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    Exclamation check derivative

    Hi could somebody please look through my answer to his homework assigment .

    Given z = sin(x + sin(t))

    show that \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}

    By using the chain-rule I get:

    f_x(x,t) = cos(x + sin(1))

    f_{xx}(x,t) = -sin(x + sin(1))

    f_t(x,t) = cos(1) \cdot cos(x + sin(1))

    f_{tt}(x,t) = 0

    Therefore

    \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}

    Does this look okay?

    Sincerely and God bless,

    Fred
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathman24 View Post
    Hi could somebody please look through my answer to his homework assigment .

    Given z = sin(x + sin(t))

    show that \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}

    By using the chain-rule I get:

    f_x(x,t) = cos(x + sin(1))

    f_{xx}(x,t) = -sin(x + sin(1))

    f_t(x,t) = cos(1) \cdot cos(x + sin(1))

    f_{tt}(x,t) = 0

    Therefore

    \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}

    Does this look okay?

    Sincerely and God bless,

    Fred
    Three things:
    1. Latex is currently down.
    2. Use instead of
    3. Your derivatives are a bit off.

    Given z = sin(x + sin(t))
    (del z)/(del x) = cos(x + sin(t)) *1

    not cos(x + sin(1)).

    This is the only error I saw, though you kept doing it consistently. So your other partials are wrong as well.

    -Dan
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  3. #3
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    Hi

    There then the partials must be

    f_{xx}(x,t) =-sin(x + sin(t))

    f_t(x,t) =     cos(t) \cdot cos(x + sin(t))

    f_tt(x,t) =    -sin(t) * cos(x+sin(t)) - (cos(t))^2 * sin(x+sint)).

    I have checked them myself, but do they look right to you now?

    Sincerley Fred

    p.s. my original reasoning regarding the chain rule, was that correct. I hope.?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathman24 View Post
    Hi

    There then the partials must be

    f_{xx}(x,t) =-sin(x + sin(t))

    f_t(x,t) =     cos(t) \cdot cos(x + sin(t))

    f_tt(x,t) =    -sin(t) * cos(x+sin(t)) - (cos(t))^2 * sin(x+sint)).

    I have checked them myself, but do they look right to you now?

    Sincerley Fred

    p.s. my original reasoning regarding the chain rule, was that correct. I hope.?
    It looks good now. In regard to your "PS" yes.

    -Dan
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