1 Answers

Best Answer

e^(z)=xyz
e^(z)-xyz=0
F(x,y,z)=e^(z)-xyz=0
According to formula 7
Suppose that an equation F(x,y,z)=0 defines z Implicitly in terms of x and y
Then
(∂ z)/(∂ x)=- (F_(x))/(F_(z)) (∂ z)/(dy)=- (F_(y))/(F_(z))
F_(x)=-yz
F_(y)=-xz
F_(z)=e^(z)-xy
Therefore
(∂ z)/(dx)=- (F_(x))/(F_(z))=- (-yz)/(e^(z)-xy)= (yz)/(e^(z)-xy)
(∂ z)/(dy)=- (F_(y))/(F_(z))=- (-xz)/(e^(z)-xy)= (xz)/(e^(z)-xy)
Result:
(∂ z)/(dx)= (yz)/(e^(z)-xy) (∂ z)/(dy)= (xz)/(e^(z)-xy)