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Math Help - differentiability of 2-variable-function

  1. #1
    Senior Member Sambit's Avatar
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    Question differentiability of 2-variable-function

    The function  f:R^2-->R is defined by <br />
f(x,y) = \frac{sin(x^2y)}{x^2+y^2} , if (x,y)\neq(0,0)
    =  0 , if (x,y)\neq(0,0)

    then the function
    (a) is differentiable at  (0,0)
    (b) is continuous at  (0,0) , but not differentiable
    (c) is not continuous at  (0,0)
    (d) has continuous partial derivatives at  (0,0)
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  2. #2
    MHF Contributor

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    Well, what have you done? Have you even made an attempt? I assume you have and it would help us to know what you tried and where you ran into problems.

    Is f continuous at (0, 0)? What are the partial derivatives? What are the partial derivatives at (0, 0)? Is there a tangent plane to the surface z= \frac{sin(x^2y)}{x^2+ y^2} at (0, 0, 0)?
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  3. #3
    Senior Member Sambit's Avatar
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    we have  f_x(a,b) = lim_{h\to0}\frac{f(a+h,b) - f(a,b)}{h}
    and  f_y(a,b) = lim_{k\to0}\frac{f(a,b+k) - f(a,b)}{k}
    therefore,  f_x(0,0) = lim_{h\to0}\frac{f(h,0) - f(0,0)}{h} = lim_{h\to0}\frac{\sin(h^20)}{h^2+0^2}\frac{1}{h} = lim_{h\to0}\frac{0}{h^3} = 0. similarly, we get  f_y(0,0) = 0.

    this is alll that i could do. don't know whether correct or not.

    how can i test its continuity at (0,0)?
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  4. #4
    MHF Contributor

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    You test continuity by seeing if it meets the definition of "continuous"!
    Is \lim_{(x,y)\to (0,0)} f(x, y)= f(0, 0)?

    That is, is \lim_{(x,y)\to (0,0)} \frac{sin(x^2y)}{x^2+ y^2}= 0?

    Try changing to polar coordinates. That way the "distance from (0, 0)" is determined by the single variable, r.

    Your partial derivatives, at (0, 0), are correct. You might also want to calculate the partial derivatives around (0, 0). There is a theorem that says that if the partial derivatives are continuous in a neighborhood of a point, then the function is differentiable at the point.
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  5. #5
    Senior Member Sambit's Avatar
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    if i take x=r\cos\theta , y=r\sin\theta, the function becomes  \frac{\sin(r^3\sin\theta\cos^2\theta)}{r^2} < \frac{1}{r^2}.
    then what?

    one thing that i observe is now if we write  \frac{1}{r^2} < \epsilon , then we we can not write  \|x-a| < \delta or  \|y-b| < \delta for some  a, b,\delta (\delta, however small.)
    does this help in any way?
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