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Linear approximation

  1. Feb 1, 2012 #1
    Find the linear approximation to the equation f(x,y) = 3 sqrt((x y)/4) at the point (2,8,6), and use it to approximate f(2.28,8.22). I know you take the derivative of fx(x,y) and fy(x,y), I think I'm taking the derivative wrong. Then after that you put x and y in the equation and solve for fx(2,8) and fy(2,8). Then take f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b). For fx(x,y) I'm coming up with .75(y/4) and for fy(x,y) I'm getting .75(x/4)
     
  2. jcsd
  3. Feb 1, 2012 #2

    jbunniii

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    Your derivatives are indeed wrong. Let's start with a simpler function of just one variable.

    [tex]g(x) = 3 \sqrt{x / 4}[/tex]


    What is [itex]g'(x)[/itex]?
     
  4. Feb 1, 2012 #3
    g(x)=3(x/4)^-.5
    g'(x)= [3(x)^-.5]/4
     
  5. Feb 1, 2012 #4
    Want to know an easy way to derive a linear approximation equation? Well, I'll give you some intuition: it's basically the same thing as the tangent line equation in so many words.
     
  6. Feb 1, 2012 #5
    L(x)= f(x) + f'(a)(x-a)

    (2.8,6) is the point...and so on
     
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