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Directional derivatives and partial derivatives

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose f: R -> R is differentiable and let h(x,y) = f(√(x^2 + y^2)) for x ≠ 0. Letting r = √(x^2 + y^2), show that:

    x(dh/dx) + y(dh/dy) = rf'(r)

    2. Relevant equations



    3. The attempt at a solution
    I have begun by showing that rf'(r) = sqrt(x^2 + y^2) * limt->0 (f(r+t) - f(r))/t

    and written out the definition form of the directional derivatives. I cant seem to find a way to equate both sides of the equation. Can anyone help?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 12, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I don't see that there is any "directional derivative" involved here. You are given that [itex]h(x,y)= f(\sqrt{x^2+ y^2})[/itex] . By the chain rule [itex]\partial f/\partial y= (df/dr)(\partial r/\partial y)[/itex] and [itex]\partial f/\partial x= (df/dr)(\partial r/\partial x[/itex].

    With [itex]r= \sqrt{x^2+ y^2}[/itex], it is easy to find [itex]\partial r/\partial x[/itex] and [itex]\partial r/\partial y[/itex].
     
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