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How to get the second derivative of this function ?

  1. Dec 31, 2008 #1
    Let a function [tex]v[/tex] in one variable, say [tex]u[/tex]
    [tex]u[/tex] is a function also, but in two variables, say [tex]x[/tex] & [tex]y[/tex]

    for the first derivative of [tex]v[/tex], i did the following:
    [tex]\frac{\partial v}{\partial x} = \frac{d v}{d u} . \frac{\partial u}{\partial x}[/tex]

    And the resutlt is:
    [tex]\frac{\partial v}{\partial x} = cos(u) . u_x[/tex]

    Note: [tex]u_x[/tex] is so hard to get, may be impossible without computer.

    But, if i want the second derivative of [tex]v[/tex], How to ?

    [tex]\frac{\partial^2 v}{\partial x^2} = ?[/tex]

    i think that the way for getting it is Differentiating the result, means differentiating: [tex]cos(u) . u_x[/tex]
    But how to also ?
     
    Last edited: Dec 31, 2008
  2. jcsd
  3. Dec 31, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi AbuYusufEg! Welcome to PF! :smile:

    Just treat cos(u) as your new v, and use the product rule.

    (there'll be a (ux)2 in it … was that putting you off? :wink:)

    Have a go! :smile:
     
  4. Dec 31, 2008 #3
    i'll give a try !, tell me if any thing wrong ..
    [tex]\frac{\partial^2 v}{\partial x^2} = -sin(u).u_x + cos(u).u_{xx}[/tex]
    is that correct ? and where is the [tex](u_x)^2[/tex] you talk about ? and why should it be here ?
     
  5. Dec 31, 2008 #4

    tiny-tim

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    The second term is correct, but the first one should be -sin(u).ux2

    can you see why? :smile:
     
  6. Dec 31, 2008 #5
    Sorry, But No !
     
  7. Dec 31, 2008 #6

    tiny-tim

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    ok … you have to use the product rule on cos(u).ux

    so it's cos(u).(∂/∂x(ux)) + (∂/∂x(cos(u)).ux

    = cos(u).uxx + … ? :smile:
     
  8. Dec 31, 2008 #7

    HallsofIvy

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    [itex]v_x= cos(u)u_x[/itex]
    use the product rule:

    [itex]v_{xx}= (cos(u))_x u_x+ cos(u) u_{xx}[/itex]
    And [itex](cos(u))_x= -sin(u) u_x[/itex] by the chain rule. Because of the "[itex]u_x[/itex]" in that,
    [tex](cos(u))_x u_x= (-sin(u) u_x)u_x= -sin(u) u_x^2[/tex].
     
  9. Dec 31, 2008 #8
    it's clear now !, good.
     
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