sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need first order derivatives first such that:

`f_x(x,y) = ((6x + 5y)(xy+1) - y(17+3x^2-2y^3+5xy))/((xy+1)^2)`

`f_x(x,y) = ((6x^2y + 6x + 5xy^2 + 5y - 17y - 3x^2y+2y^4-5xy^2))/((xy+1)^2)`

`f_x(x,y) = (3x^2y + 6x- 12y + 2y^4)/((xy+1)^2)`

`f_y(x,y) = ((-6y^2 + 5x)(xy+1) - x(17+3x^2-2y^3+5xy))/((xy+1)^2)`

`f_y(x,y) = (-6xy^3 - 6y^2 + 5x^2y + 5x - 17x - 3x^3 + 2xy^3-5x^2y)/((xy+1)^2)`

`f_y(x,y) = (-4xy^3 - 6y^2 - 12x - 3x^3)/((xy+1)^2)`

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

`f_(x x) = ((6xy + 6)(xy+1)^2 - 2y(xy+1)(3x^2y + 6x- 12y + 2y^4))/((xy+1)^4)`

`f_(x x) = ((6xy + 6)(xy+1) - 2y(3x^2y + 6x- 12y + 2y^4))/((xy+1)^3)`

`f_(x x) = (6x^2y^2 + 12xy + 6 - 6x^2y^2 - 12xy + 24y^2 - 4y^5)/((xy+1)^3)`

`f_(x x) = (6+ 24y^2 - 4y^5)/((xy+1)^3)`

`f_(xy) = ((3x^2-12+8y^3)((xy+1)^2) - 2x(xy+1)(3x^2y + 6x- 12y + 2y^4))/((xy+1)^4)`

`f_(xy) = ((3x^2-12+8y^3)(xy+1) - 2x(3x^2y + 6x- 12y + 2y^4))/((xy+1)^3)`

`f_(xy) = (3x^3y + 3x^2 y^4 + 8y^3 - 6x^3y - 12x^2 + 24xy - 4xy^4)/((xy+1)^3)`

`f_(yy) = ((-12xy^2 - 12y)((xy+1)^2) - 2x(xy+1)(-4xy^3 - 6y^2 - 12x - 3x^3))/((xy+1)^4)`

`f_(yy) = ((-12xy^2 - 12y)(xy+1) - 2x(-4xy^3 - 6y^2 - 12x - 3x^3))/((xy+1)^3)`

`f_(yy) = (-12x^2y^3 -24xy^2 - 12y + 8x^2y^3+ 12xy^2+ 24x^2 +6x^4)/((xy+1)^3)`

`f_(yy) = (-4x^2y^3 -12xy^2 - 12y + 24x^2 + 6x^4)/((xy+1)^3)`

`f_(yx) = ((-4y^3 - 12 - 9x^2)((xy+1)^2) - 2y(xy+1)(-4xy^3 - 6y^2 - 12x - 3x^3))/((xy+1)^4)`

`f_(yx) = ((-4y^3 - 12 - 9x^2)(xy+1) - 2y(-4xy^3 - 6y^2 - 12x - 3x^3))/((xy+1)^3)`

`f_(yx) = (-4xy^4 - 4y^3 - 12xy - 12 - 9x^3y - 9x^2+8xy^4+ 12y^3+ 24xy+ 6x^3y))/((xy+1)^3)`

`f_(yx) = (4xy^4 - 12 - 3x^3y - 9x^2 + 8y^3 + 12xy)/((xy+1)^3)`

Hence, evaluating the second order partial derivatives yields `f_(x x) = (6+ 24y^2 - 4y^5)/((xy+1)^3); f_(xy) = (3x^3y + 3x^2 y^4 + 8y^3 - 6x^3y - 12x^2 + 24xy - 4xy^4))/((xy+1)^3) ; f_(yy) = (-4x^2y^3 -12xy^2 - 12y + 24x^2 + 6x^4))/((xy+1)^3) ; f_(yx) = (4xy^4 - 12 - 3x^3y - 9x^2 + 8y^3 + 12xy))/((xy+1)^3).`