Alle partiellen Ableitungen erster u. zweiter Ordnung von f(x,y,z)=ln(√(x^2 + y^2 + z^2)) und f(x,y)=(x+y)/(x^2 + y^2 )

Question

Berechnen Sie alle partiellen Ableitungen erster und zweiter  Ordnung der folgenden Funktion:

f(x, y, z) = ln(√(x² + y² + z²))

f(x, y) = (x + y)/(x² + y²)

Answer 1

f(x, y, z) = ln(√(x^2 + y^2 + z^2))

df/dx = x/(x^2 + y^2 + z^2)
df/dy = y/(x^2 + y^2 + z^2)
df/dz = z/(x^2 + y^2 + z^2)

d^{2}f/dxdx = (- x^2 + y^2 + z^2)/(x^2 + y^2 + z^2)^2
d^{2}f/dxdy = - 2·x·y/(x^2 + y^2 + z^2)^2
d^{2}f/dxdz = - 2·x·z/(x^2 + y^2 + z^2)^2

d^{2}f/dydx = - 2·x·y/(x^2 + y^2 + z^2)^2
d^{2}f/dydy = (x^2 - y^2 + z^2)/(x^2 + y^2 + z^2)^2
d^{2}f/dydz = - 2·y·z/(x^2 + y^2 + z^2)^2

d^{2}f/dzdx = - 2·x·z/(x^2 + y^2 + z^2)^2
d^{2}f/dzdy = - 2·y·z/(x^2 + y^2 + z^2)^2
d^{2}f/dzdz = (x^2 + y^2 - z^2)/(x^2 + y^2 + z^2)^2

Comments

f(x, y) = (x + y)/(x^2 + y^2)

df/dx = (- x^2 - 2·x·y + y^2)/(x^2 + y^2)^2
df/dy = (x^2 - 2·x·y - y^2)/(x^2 + y^2)^2

d^{2}f/dxdx = 2·(x^3 + 3·x^2·y - 3·x·y^2 - y^3)/(x^2 + y^2)^3
d^{2}f/dxdy = 2·(- x^3 + 3·x^2·y + 3·x·y^2 - y^3)/(x^2 + y^2)^3

d^{2}f/dydx = 2·(- x^3 + 3·x^2·y + 3·x·y^2 - y^3)/(x^2 + y^2)^3
d^{2}f/dydy = 2·(- x^3 - 3·x^2·y + 3·x·y^2 + y^3)/(x^2 + y^2)^3