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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`

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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
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