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Math Help - Partial Derivatives

  1. #1
    Senior Member polymerase's Avatar
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    Partial Derivatives

    Let f(x,y)=x^2\arctan\frac{x}{y}-y^2\arctan\frac{x}{y} if xy\neq0
    Let f(x,y)=0 if xy=0

    Compute \frac{\partial}{\partial y}\frac{\partial f}{\partial x}(0,0) and \frac{\partial}{\partial x}\frac{\partial f}{\partial y}(0,0) if they exist.

    Ive proven that f(x,y) is continuous at 0, but I'm not sure which function I should be using to calculate the derivative, any help?

    Thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by polymerase View Post
    Let f(x,y)=x^2\arctan\frac{x}{y}-y^2\arctan\frac{x}{y} if xy\neq0
    Let f(x,y)=0 if xy=0

    Compute \frac{\partial}{\partial y}\frac{\partial f}{\partial x}(0,0) and \frac{\partial}{\partial x}\frac{\partial f}{\partial y}(0,0) if they exist.

    Ive proven that f(x,y) is continuous at 0, but I'm not sure which function I should be using to calculate the derivative, any help?

    Thanks
    Start by computing that f_{xy} and f_{yx} and then check as to whether or not they are continous at x,y=0. If so then thats your answer. Note based on these conditions that f_{xy}=f_{yx}.
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  3. #3
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    Quote Originally Posted by polymerase View Post
    Let f(x,y)=x^2\arctan\frac{x}{y}-y^2\arctan\frac{x}{y} if xy\neq0
    Let f(x,y)=0 if xy=0

    Compute \frac{\partial}{\partial y}\frac{\partial f}{\partial x}(0,0) and \frac{\partial}{\partial x}\frac{\partial f}{\partial y}(0,0) if they exist. (I'll denote these derivatives by \color{blue}f_{xy} and \color{blue}f_{yx} respectively.)

    Ive proven that f(x,y) is continuous at 0, but I'm not sure which function I should be using to calculate the derivative, any help?
    For a function where the defining formula breaks down at some points, you have to go back to the definition of the derivative at those points.

    For the function given above, you can use the usual rules of differentiation to find the partial derivatives of (x^2-y^2)\arctan\tfrac{x}{y} at points where neither x nor y is zero. But if you want to find for example f_x(0,y) (where y≠0) then you have to evaluate the limit given by the definition f_x(0,y) = \lim_{h\to0}\frac{f(h,y)-f(0,y)}h = \lim_{h\to0}\frac{(h^2-y^2)\arctan\frac hy-0}h. To find this limit, you'll need to use the fact that \lim_{x\to0}\frac{\arctan x}x=1. Leaving out the details, this gives f_x(0,y) = -y.

    To find f_x(0,0), you again have to go back to the definition of the derivative, but this time it's easier: f_x(0,0) = \lim_{h\to0}\frac{f(h,0)-f(0,0)}h = \lim_{h\to0}\frac{0-0}h = 0. So in fact the formula f_x(0,y) = -y still works when y=0.

    To calculate the second partial derivative f_{xy}(0,0) you again have to go back to the definition: f_{xy}(0,0) = \lim_{k\to0}\frac{f_x(0,k)-f_x(0,0)}k = \lim_{k\to0}\frac{-k-0}k = -1.

    For the other mixed partial derivative f_{yx}(0,0), *if it exists*, you need to use the same procedure, but this time finding f_y(x,0) and f_y(0,0) first, and then using f_{yx}(0,0) = \lim_{h\to0}\frac{f_y(h,0) - f_y(0,0)}h.
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