Partial Derivatives

Question

Let the surface z = f(x,y) be defined implicitly by the equation: sin(xyz)=x^(2)y^(2)+ z^(2) − 1 .

(a). Find the partial derivatives ∂z ∂x and ∂z ∂y.
(b). Find the equation of the tangent plane at the point (1,1,0).

Answer 1

sin(xyz)=x²y² + z² - 1

 

cos(xyz) (yz + xy ∂z/∂x) = 2xy² + 2z ∂z/∂x

 

[xy cos(xyz) - 2z] ∂z/∂x = 2xy² - yz cos xyz

 

∂z/∂x = [2xy² - yz cos(xyz)]/[xy cos(xyz) - 2z]

 

By symmetry:

 

∂z/∂y = [2x²y - xz cos(xyz)]/[xy cos(xyz) - 2z]

 

(∂z/∂x)(1,1,0) = (∂z/∂y)(1,1,0) = 2

 

The tangent plane at (1,1,0) is:

 

z = 2x + 2y - 4