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Homework Help: Lost differentiating a function

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data The problem statement, all variables and given/known data[/b]
    Hello. I'm trying to solve a non-linear problem and I have been working through the notes on this pdf to try and understand the method before I use it but I get stuck at one of the steps. The pdf is here:
    I cannot follow how the author went from equation 9.7.8 to 9.7.9

    2. Relevant equations
    The author has the following equation:

    [tex]g (\lambda) \equiv f(xold + \lambda p) [/tex]
    He differentiates this function with respect to [tex] \lambda [/tex] and gets the following
    [tex]g'(\lambda) = \nabla f \cdot p[/tex]

    3. The attempt at a solution
    My understanding is that the author differentiated with respect to [tex] \lambda[/tex] so they got the [tex]\nabla f[/tex] and then differentiated what was inside the function to get the [tex]p[/tex] value. But I don't see how it became a dot product, or am I just misreading it?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 2, 2009 #2
    Welcome to PF, o_damhsa

    The pdf doesn't open for me, but here's how it looks from what you've said.

    Call [tex]xold=\langle x_1,x_2,x_3\rangle[/tex] and [tex]p=\langle p_1,p_2,p_3 \rangle[/tex]

    Then [tex]g(\lambda)=f(x_1+\lambda p_1,x_2+\lambda p_2,x_3+\lambda p_3)[/tex]

    Now by the Chain Rule,

    \frac{\partial f}{\partial x}\frac{dx}{d\lambda}
    +\frac{\partial f}{\partial y}\frac{dy}{d\lambda}
    +\frac{\partial f}{\partial z}\frac{dz}{d\lambda} [/tex]

    \frac{\partial f}{\partial x} p_1
    +\frac{\partial f}{\partial y} p_2
    +\frac{\partial f}{\partial z} p_3[/tex]

    \left\langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right\rangle\cdot\langle p_1,p_2,p_3 \rangle [/tex]

    \nabla f\cdot p [/tex]
  4. Oct 5, 2009 #3
    Hello Billy Bob,

    Thank you very much for your reply! It makes perfect sense now :smile:

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