Partial Derivative Help?

Question

If sin( −1x − 2y + z) = 0, find the first partial
derivatives ∂z/∂x and ∂z/∂y at the point (0, 0, 0).
A. ∂z/∂x (0, 0, 0) =
B. ∂z/∂y (0, 0, 0) =

Answer 1

If sin( −1x − 2y + z) = 0, then −1x − 2y + z = sin .inv (0)
-x-2y+z = 0
z = x+2y

A. ∂z/∂x (0, 0, 0) = 1
B. ∂z/∂y (0, 0, 0) = 2

Answer 2

-x - 2y + z = 0 or nπ
The RHS doesn't really matter
z = x + 2y + C

differentiating
∂z/∂x = 1
∂z/∂y = 2

This is constant at all points.

Answer 3

Question A
Assume question could read as:-
f (x,y,z) = sin(-x - 2y + z)
ðf/ðx = (-1).cos(- x - 2y + z)
ðf/ðx = (-1) at (0,0,0)

Question B
ðf/ðy = (-2).cos(- x - 2y + z)
ðf/ðy = (-2) at (0,0,0)