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Differentiability of a function

  1. Aug 15, 2012 #1
    Hello,

    I'm having problems figuring out theoretical problem on "differentiability of a function". [I hope that I spelled it right...]

    Suppose that:
    1. Functions f(x,y) and g(x,y) are well defined in some little domain around (0,0). (*1)
    2. g(x,y) is continuous at (0,0). (*2)
    3. f(x,y) is differentiable and continuous at (0,0) and f(0,0)=0. (*2)

    Prove that h(x,y)=f(x,y)g(x,y) is differentiable at (0,0).

    What I tried:
    Let ε>0
    From (*1) and (*2) I know that for every ||(x,y)||<δ_1 |g(x,y)-g(0,0)|<ε/2. (*4)
    From (*1) and (*3) I know that there are a,b and u(x,y), v(x,y) so that:
    f(x,y)=f(0,0)+ax+by+xu(x,u)+yv(x,u) ; u(x,y) and v(x,y) approaches 0 when (x,y) approaches (0,0). (*5)

    So from (*4) and (*5) and a little of algebra I get that:
    [tex]
    x[g(0,0)-\epsilon/2]+y[g(0,0)-\epsilon/2]+xg(x,y)v(x,y)+yg(x,y)v(x,y) \leq h(x,y) \leq x[g(0,0)+\epsilon/2]+y[g(0,0)+\epsilon/2]+xg(x,y)v(x,y)+yg(x,y)v(x,y)
    [/tex]

    And at this point I have no idea what to do and I fear my approach to this problem is wrong...
     
  2. jcsd
  3. Aug 15, 2012 #2
    If f(x, y) is differentiable at (0, 0), there exists a linear map [itex]df_{(0, 0)}(u, v)[/itex] such that [tex]\lim_{(u, v) \rightarrow (0, 0)} \frac {f(u, v) - df_{(0, 0)}(u, v) - f(0, 0)} {||(u, v)||} = \lim_{(u, v) \rightarrow (0, 0)} \frac {f(u, v) - df_{(0, 0)}(u, v)} {||(u, v)||} = 0[/tex] Now check whether [itex]g(0, 0) \cdot df_{(0, 0)}(u, v)[/itex] is such a linear map for h(x, y) at (0, 0).
     
  4. Aug 15, 2012 #3
    I'm sorry but I'm still not getting it, ofcourse I tried to find a, b so that:

    [tex]\lim_{(x,y)\rightarrow(0,0)} \frac {h(x,y)-h(0,0)-ax-by}{||(x,y)||}=0[/tex]
    but wasnt able to find appropriate a,b...
     
    Last edited: Aug 15, 2012
  5. Aug 15, 2012 #4
    [itex]g(0, 0) \cdot df_{(0, 0)}(x, y) = ax + by[/itex] is what you are looking for. You need to show that.
     
  6. Aug 15, 2012 #5
    I appreciate your help, however your hints are points that I already know. [I think you get the wrong impression that I'm not familiar with the basic staff, indeed a lot of time passed since I took that course however I do remember the basics (I'm retaking the exam to improve my GPA)]

    I am familiar with the different criterias for differentiability and I know that if I can show that:
    [tex]\lim_{(x,y)\rightarrow (0,0)} \frac{f(x,y)g(x,y)-xf_x(0,0)g(0,0)-yf_y(0,0)g(0,0)}{\sqrt{x^2+y^2}}=0 [/tex] then I will solve the problem but the problem is that I COULDN"T show that. [this was the first thing I tried]

    There is also another criteria [which I remember I was using with great success] for differentiability: there are scalars a,b [in my case a=f_x, b=f_y] and functions u(x,y),v(x,y) [both functions approaches to 0 when (x,y) goes to (0,0)] so that:
    [tex]f(x,y)=f(0,0)+xf_x(0,0)+yf_y(0,0)+xu(x,y)+yv(x,y)[/tex]
    Now if I multiply [and remember f(0,0)=0] both sides by g(x,y) I get:
    [tex]f(x,y)g(x,y)=xf_x(0,0)g(x,y)+yf_y(0,0)g(x,y)+xu(x,y)g(x,y)+yv(x,y)g(x,y)[/tex] and this is also very close to what I need:
    [tex]f(x,y)g(x,y)=x[f_x(0,0)g(0,0)]+y[f_y(0,0)g(0,0)]+x[u(x,y)g(x,y)]+y[v(x,y)g(x,y)][/tex]
    where "the new a" is f_x(0,0)g(0,0) "the new b" is [f_y(0,0)g(0,0)] "the new u(x,y)" is [u(x,y)g(x,y)] and so on, I'm close to this but I'm not there...

    What I tried to do in my first post is:
    I arrived to:
    [tex]f(x,y)g(x,y)=xf_x(0,0)g(x,y)+yf_y(0,0)g(x,y)+xu(x,y)g(x,y)+yv(x,y)g(x,y)[/tex]
    So due to continuity of g I can conclude:
    [tex]xf_x(0,0)[g(0,0)+\epsilon/2]+yf_y(0,0)[g(0,0)+\epsilon/2]+xu(x,y)g(x,y)+yv(x,y)g(x,y) \leq h(x,y) \leq xf_x(0,0)[g(0,0)+\epsilon/2]+yf_y(0,0)[g(0,0)+\epsilon/2]+xu(x,y)g(x,y)+yv(x,y)g(x,y)[/tex]
    and it is very tempting to say the following [which is most likely not "legal"]: because this inequality is true for every epsilon that I can "swap" it to:
    h(x,y) = xf_x(0,0)[g(0,0)]+yf_y(0,0)[g(0,0)]+xu(x,y)g(x,y)+yv(x,y)g(x,y)[/tex]

    Sorry if this post is disorganized, I tried to think out loud...
     
    Last edited: Aug 15, 2012
  7. Aug 16, 2012 #6
    This is not correct. It should be [tex]f(x,y)=f(0,0)+xf_x(0,0)+yf_y(0,0)+R(x,y)[/tex] where [tex]\lim_{(x, y) \rightarrow (0, 0)} \frac {R(x, y)} {||(x, y)||} = 0[/tex] That is, R need not be linear but it must go to zero faster than the norm of the point.

    Then [tex]f(x,y)g(x, y) = f(0,0)g(x, y) + [xf_x(0,0) + yf_y(0,0)]g(x, y) + g(x, y)R(x,y)[/tex][tex] = f(0,0)g(x, y) + [xf_x(0,0) + yf_y(0,0)]g(0, 0) + [xf_x(0,0) + yf_y(0,0)]g(x, y) - [xf_x(0,0) + yf_y(0,0)]g(0, 0) + g(x, y)R(x,y)[/tex][tex] = f(0,0)g(x, y) + [xf_x(0,0) + yf_y(0,0)]g(0, 0) + [xf_x(0,0) + yf_y(0,0)] \cdot [g(x, y) - g(0, 0)] + g(x, y)R(x,y)[/tex][tex] = f(0,0)g(x, y) + [xf_x(0,0) + yf_y(0,0)]g(0, 0) + R'(x, y)[/tex] Prove [tex]\lim_{(x, y) \rightarrow (0, 0)} \frac {R'(x, y)} {||(x, y)||} = 0[/tex]
     
  8. Aug 16, 2012 #7
    Are you sure about that?
    I have a theorem in my book that proves that the following 2 are "equivalent".
    1. [itex] f(x,y)=o(||(x,y)||)[/itex] when [itex](x,y) \rightarrow (0,0)[/itex]
    2. there are functions u(x,y) and v(x,y) defined around (0,0) so that in some domain around this point the following holds:
    [itex]f(x,y)=xu(x,y)+yv(x,y) [/itex] when u,v goes to 0 around (0,0).

    And I think these 2 are the same [equivalent] to what you have written.

    In this same theorem the book proves some different criteria for differentiability. [unfortunately it somewhat long to copy to the forum and written in Hebrew]

    And btw your last hint, was very helpful I guess I can prove what is needed now, however I hope you can help me clear the above point.
    Again thank you very much.
     
    Last edited: Aug 16, 2012
  9. Aug 16, 2012 #8
    What I wrote is the definition of differentiability. This is all you really need here.

    What you wrote is a more complex expression having the same effect (which seems true), but it just makes things seem, well, more complex.
     
  10. Aug 16, 2012 #9
    I see.
    A million thanks for your help. [This trick with adding something and then subtracting it, I always forget about it...]
     
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