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Prove differentiable for a function

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data
    f(x,y) is differentiable in (0,0) and f(0,0)=0
    q(t) devirate t=0 and q'(0)=1 q(0)=0
    let it be g(x,y) = q(f(x,y))
    prove that g differentiable in (0,0) and that [tex]f_{x}(0,0) = g_{x}(0,0)[/tex]


    2. Relevant equations
    all calculus

    3. The attempt at a solution
    Well my idea is like this
    First I know that because f(x,y) differentiable at (0,0) so
    [tex]f(0+x,0+y)= f(0,0) +f_{x}(0,0)*x + f_{y}(0,0)*y + o(||x,y||)[/tex]

    [tex] lim_{(x,y)->(0,0)} \frac{f(x,y)-f_{x}(0,0)*x - f_{y}(0,0)*y}{||x,y||} = 0[/tex]

    and when trying to prove that g(x,y) is differentiable I need to prove that
    [tex]g(0+x,0+y) = g(0,0) + g_{x}(0,0)*x + g_{y}(0,0)*y + o(||x,y||)[/tex]
    [tex]g(0+x,0+y) = 0 + f_{x}(0,0)*q'(0)*x + f_{y}(0,0)*q'(0)*y + o(||x,y||)[/tex]

    [tex]g(0+x,0+y) = 0 + f_{x}(0,0)*1*x + f_{y}(0,0)*1*y + o(||x,y||)[/tex]
    [tex] lim_{(x,y)->(0,0)} \frac{q(f(x,y))-f_{x}(0,0)*x - f_{y}(0,0)*y}{||x,y||} = 0[/tex]
    And I'm a bit stuck in here..
    logically I think I can use the fact that q'(0) = 1
    [tex] lim_{(x)->(0)} \frac{q(0+x)+q(0)}{x} =\frac{q(x)}{x}= 1[/tex]
    But I dont manage to figure how

    Thank you
     
    Last edited: Apr 13, 2010
  2. jcsd
  3. Apr 14, 2010 #2
    Anyone?
    I've made some more progress but its not enough,see
    Because I know that q(x) has a derivative in t=0 then it means that
    q(0+x) = q(0)+q'(0)x+o(|x|) (x->0)
    Thats mean that: [ q(0)= 0 q'(0)=1)
    [tex]lim_{x->0}\frac{q(x)-x}{|x|}=0[/tex]

    And I think maybe I need to use this with the knowledge that

    [tex] lim_{(x,y)->(0,0)} \frac{f(x,y)-f_{x}(0,0)*x - f_{y}(0,0)*y}{||x,y||} = 0[/tex]

    Meaning that
    [tex]f(x,y)-f_{x}(0,0)*x - f_{y}(0,0)*y = v(x,y)\sqrt{x^{2}+y^{2}}[/tex]
    while [tex]lim_{(x.y)->(0,0)}v(x,y)=0[/tex]
    And that
    q(f(x,y)) = g(x,y)

    But I dont know how

    Anyone Please?
     
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