Question

If z = f (x, y) where x = e^u +e^-v, y = e^-u - e^v then prove that `(delz)/(delu)-(delz)/(delv)=x(delz)/(delx)-y(delz)/(dely).`

Answer 1

Given: z = f (𝑥, y) , x = e^u +e^-v ……… (1)
y = e^-u - e^v ………… (2)
By Chain Rule,

`(delz)/(delu)-(delzdelx)/(delvdelu)=(delzdely)/(delydelu)`……… (3)

And 

`(delz)/(delv)-(delzdelx)/(delxdelv)=(delzdely)/(delydelv)`……… (3)

∴From equation 1 and 2,

`(delx)/(delu)=e^U`   `(delx)/(delv)=-e^(-V)`

`(dely)/(delv)=-e^(-U)`    `(dely)/(delv)=-e^(V)`

∴ From equation 3 and 4,

`(delz)/(delu)=e^U (delz)/(delx)- e^(-U)(delz)/(dely)`………. (5)

And

`(delz)/(delv)=e^(-V) (delz)/(delx)- e^V(delz)/(dely)`………. (6)

By Subtracting Equation 5 and 6,

`(delz)/(delu)-(delz)/(delv)=(e^U+e^(-V))(delz)/(delx)-(e^(-U)-e^V)(delz)/(dely)`

=` x (delz)/(delx)y(delz)/(dely)`……………………. (By using equation 1 and 2)

`(delz)/(delu)-(delz)/(delv)=x(delz)/(delx)-y(delz)/(dely)`

Hence proved.