A function f(x, y) is called continuously differentiable at a point (x_0, y_.0)

Question

A function f(x, y) is called continuously differentiable at a point (x_0, y_.0) if both of its partial derivatives exist and are continuous at (x_0, y_0). We also call them C^1 functions, i.e., f being a C^1 function means f(x, y) is continuously differentiable. Decide whether the following functions are C^1 at all points (x, y) elementof R^2: (a) f(x, y) = {0, if (x, y) = (0, 0), 2xy/(x^2 + y^2)^2, if (x, y) notequalto (0, 0). (b) f(x) = {0, if (x, y) = (0, 0), xy/squareroot x^2 + y^2, if (x, y) notequalto (0, 0).A function f(x, y) is called continuously differentiable at a point (x_0, y_.0) if both of its partial derivatives exist and are continuous at (x_0, y_0). We also call them C^1 functions, i.e., f being a C^1 function means f(x, y) is continuously differentiable. Decide whether the following functions are C^1 at all points (x, y) elementof R^2: (a) f(x, y) = {0, if (x, y) = (0, 0), 2xy/(x^2 + y^2)^2, if (x, y) notequalto (0, 0). (b) f(x) = {0, if (x, y) = (0, 0), xy/squareroot x^2 + y^2, if (x, y) notequalto (0, 0).

Explanation / Answer

a> f(x,y) = 2xy/(x^2+y^2)^2 , if (x,y) not equal to 0

=> fx = [ 2y(x^2+y^2)^2 - 2xy(2(x^2+y^2)*2x) ]/(x^2+y^2)^4

fx = [ 2y(x^2+y^2)^2 - 8x^2y(x^2+y^2)]/(x^2+y^2)^4

fx = [2y(x^2+y^2) - 8x^2y]/(x^2+y^2)^3

fx = [2y^3 - 6x^2y]/ (x^2+y^2)^3

fy = [ 2x(x^2+y^2)^2 - 2xy(2(x^2+y^2)*2y) ]/(x^2+y^2)^4

fy = [ 2x(x^2+y^2)^2 - 8xy^2(x^2+y^2)]/(x^2+y^2)^4

fy = [2x^3 - 6xy^2]/ (x^2+y^2)^3

and when (x,y) = (0,0)

fx = fy = 0

the partial derivatives show that the piece wise function is continuous.

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