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How do I find the directional derivative?

  1. Jul 13, 2008 #1
    Find the directional derivative of f(x,y)= ln sqrt(x^2+y^2) at (x,y) does not = (0,0) toward the origin

    I found the gradient: grad f(x,y)=1/(x^2+y^2)*(xi+yj) but I have no idea what to do after that.
  2. jcsd
  3. Jul 13, 2008 #2


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    How are you given the direction? The directional derivative of f(x,y) in the direction of the unit vector [itex]\vec{u}[/itex] is [itex]\nabla f\cdot \vec{u}[/itex].

    The directional derivative of of f(x,y) in the direction of any vector [itex]\vec{v}[/itex] can be done by finding a unit vector in that direction first: [itex]\vec{v}/|\vec{v}|[/itex] and is [itex]\nabla f\cdot \vec{v}/|\vec{v}|[/itex].

    If the direction is at angle [itex]\theta[/itex] with the x-axis (whether the point at which you are finding the derivative is (0,0) or not), then a unit vector in that direction is [itex]cos(\theta)\vec{i}+ sin(\theta)\vec{j}[/itex] and so the directional derivative is
    [tex]\nabla f(x,y)\cdot (cos(\theta)\vec{i}+ sin(\theta)\vec{j})= \frac{\partial f}{\partial x}cos(\theta)+ \frac{\partial f}{\partial y}sin(\theta)[/tex]

    Of course, if the angle [itex]\theta[/itex] is with the y-axis, just swap "sine" and "cosine".
  4. Jul 13, 2008 #3
    When re-examining the problem I solved for grad f*u/|u| which gives (xi+yj)/[sqrt(x^2+y^2)]*[1/(x^2+y^2)*(xi+yj)] which when solved gives: (x^2+y^2)^(-1/2) which I believe is the correct answer.
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