# If U = E X Y Z F ( X Y Z ) Where F ( X Y Z ) is an Arbitrary Function of X Y Z . Prove That: X ∂ U ∂ X + Z ∂ U ∂ Z = Y ∂ U ∂ Y + Z ∂ U ∂ Z = 2 X Y Z . U - Applied Mathematics 1

Sum

If u=e^(xyz)f((xy)/z) where f((xy)/z) is an arbitrary function of (xy)/z.

Prove that: x(delu)/(delx)+z(delu)/(delz)=y(delu)/(dely)+z(delu)/(delz)=2xyz.u

#### Solution

Let   (xy)/z=w    therefore u=e^(xyz)f(w)

Diff. u w.r.t. x partially,

(delu)/(delx)=e^(xyz)f'(w)+f(w).e^(xyz).yz

Diff. u w.r.t y partially ,

(delu)/(dely)=e^(xyz)f'(w)+f(w).e^(xyz).yz

Diff. u w.r.t y partially,

(delu)/(delz)=e^(xyz)f'(w)+f(w).e^(xyz).xy

x(delu)/(delx)+z(delu)/(delz)=xe^(xyz)f'(w)+f(w).e^(xyz).xyz+ze^(xyz)f'(w)+f(w).e^(xyz).xyz...(1)

y(delu)/(dely)+z(delu)/(delz)=ye^(xyz)f'(w)+f(w).e^(xyz).xyz+ze^(xyz)f'(w)+f(w).e^(xyz).xyz...(2)

From (1) and (2),

x(delu)/(delx)+z(delu)/(delz)=y(delu)/(dely)+z(delu)/(delz)=2xyz.u

Hence Proved.

Concept: Partial Derivatives of First and Higher Order
Is there an error in this question or solution?