# for the function compute the second-order partial derivatives fxx, fyy, fxy and fyx of f(x,y)= (5x^2 -y)^.

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You need to evaluate the first order partial derivatives for the function `f(x,y) = (5x^2 -y)^3` such that:

`f_x = 3(5x^2 -y)^2*(del(5x^2 -y))/(del x)`

Notice that you need to differentiate with respect to x, considering y as a constant such that:

`f_x = 3(5x^2 -y)^2*(10x) => f_x = 30x*(5x^2 -y)^2`

You need to evaluate `f_y` , hence, you need to differentiate with respect to y, considering x as a constant such that:

`f_y = 3(5x^2 -y)^2*(del(5x^2 -y))/(del y)`

`f_y = -3(5x^2 -y)^2`

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

`f_(x x) = (del(30x*(5x^2 -y)^2))/(del x)`

`f_(x x) = 30*(5x^2 -y)^2 + 30x*2*(5x^2 -y)*(10x)`

`f_(x x) = 30*(5x^2 -y)^2 + 600x(5x^2 -y)`

Factoring out `30(5x^2 -y)` yields:

`f_(x x) = 30(5x^2 -y)(5x^2 -y + 20x)`

`f_(y y)= (del(-3(5x^2 -y)^2))/(del y)`

`f_(y y) = 6(5x^2 -y)`

You need to evaluate partial mixed derivatives such that:

`f_(y x) = (del f_y)/(del x)`

`f_(y x) = (del(-3(5x^2 -y)^2))/(del x)`

`f_(y x) = -6(5x^2 -y)*(10x)`

`f_(y x) = -60x*(5x^2 -y)`

`f_(x y) = (del f_x)/(del y)`

`f_(x y) = (del(30x*(5x^2 -y)^2))/(del y)`

`f_(x y) = -60x*(5x^2 -y)`

Notice that `f_(x y) = f_(y x) = -60x*(5x^2 -y)`

Hence, evaluating the first and second partial derivatives yields f`_x = 30x*(5x^2 -y)^2 ; f_y = -3(5x^2 -y)^2 ; f_(x x) = 30(5x^2 -y)(5x^2 -y + 20x) ; f_(y y) = 6(5x^2 -y) ; f_(x y) = f_(y x) = -60x*(5x^2 -y).`