Partial derivatives

Question

If u=log(x^3+y^3+z^3) then show that partial double derivative wrt x,y,z =-3/(x+y+z)^2

Answer 1

Ux = (3x^2)/(x^3 +y^3+z^3)

Uxx = {(x^3 + y^3 + z^3)(6x) - (3x^2)(3x^2)} / (x^3 + y^3 = z^3)^2

= { 6x^4 + 6xy^3 + 6xz^3 - 9x^4) / (x^3 + y^3 = z^3)^2

Uxy = (- 9x^2 y^2) / (x^3 + y^3 + z^3)^2

Uxz = (- 9x^2 z^2) / (x^3 + y^3 +z^3)^2

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Uy = 3y^2 / (x^3 + y^3 + z^3)

Uyx = (-9x^2y^2)/(x^3 + y^3 + z^3)^2 <---- notice it is equal to Uxy, which holds by theorem if continuous

Uyy = ((x^3 + y^3 + z^3)(6y) - (3y^2)(3y^2))/ (x^3 + y^3 + z^3)^2

= (6x^3y + 6y^4 + 6yz^3 - 9y^4)/(x^3 + y^3 + z^3)^2

Uyz = - (3y^2)(3z^2) / (x^3 + y^3 + z^3)^2 =

-9y^2z^2 / (x^3 + y^3 + z^3)^2

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Uz = (3z^2)/(x^3 +y^3+z^3)

Uzx = - (9z^2)/(x^3 +y^3+z^3)^2 <--- notice it is the same as Uxz

Uzy = (-9y^2z^2)/(x^3 +y^3+z^3)^2 <--- same as Uyz

Uzz = (6x^3z + 6y^3z + 6z^4-9z^4)/(x^3 +y^3+z^3)^2