Question: First describe the domain of the given function then find the partial derivatives fx, fy, fxx and fyx  f(x,y)= 2x+y/x-y

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The domain of the function consists of values of variables that make the function possible, hence, since the equation of the function is a fraction, you need to put the following condition, such that:

`x-y != 0 => x!=y`

Hence, the domain of the function consists of all real values.

You need to find the first order partial derivatives such that:

`f_x(x,y) = (del((2x+y)/(x-y)))/(del x)`

`f_x(x,y) = ((del(2x+y))/(del x)(x-y) - (2x+y)(del(x-y))/(del x))/((x-y)^2)`

`f_x(x,y) = (2(x-y) - (2x+y))/((x-y)^2)`

`f_x(x,y) = (2x - 2y - 2x - y)/((x-y)^2)`

`f_x(x,y) = (-3y)/((x-y)^2)`

`f_y(x,y) = ((del(2x+y))/(del y)(x-y) - (2x+y)(del(x-y))/(del y))/((x-y)^2)`

`f_y(x,y) = (x - y+2x+y)/((x-y)^2)`

`f_y(x,y) = (3x)/((x-y)^2)`

You need to evaluate the second order partial derivatives such that:

`f_(x x)(x,y) = ((3y)(del(x-y)^2)/(del x))/((x-y)^4) `

`f_(x x)(x,y) = ((6y)(x-y))/((x-y)^4)`

`f_(x x)(x,y) = (6y)/((x-y)^3)`

`f_(yx)(x,y) = (3x((x-y)^2) - (6x(x-y)))/((x-y)^4)`

`f_(yx)(x,y) = (3x(x-y)(x - y- 2))/((x-y)^4)`

`f_(yx)(x,y) = (3x(x - y - 2))/((x-y)^4) `

Hence, evaluating the first order and the second order partial derivatives yields `f_x(x,y) = (-3y)/((x-y)^2) ; f_y(x,y) = (3x)/((x-y)^2) ; f_(x x)(x,y) = (6y)/((x-y)^3) ; f_(yx)(x,y) = (3x(x - y - 2))/((x-y)^4).`

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