Finding $\frac{\partial z}{\partial x}$ when sin(5x-4y+z)=0

[UrbanXrisis]

sin(5x-4y+z)=0

how do I find $$\frac{\partial z}{\partial x}$$?

if the problem is sin(5x-4y+z)=f(x,y,z), I can find $$\frac{\partial f}{\partial x}$$ but I don't know what to do when it is just equal to zero.

 

[dicerandom]

Solve for z, take the partial derivative.

 

[HallsofIvy]

Simpler: take the partial derivative with respect to x, assuming that y is independent of x, z a function of x, then solve for z_x. Use the chain rule. Remember "implicit differentiation" from Calculus I?
(sin(5x- 4y+ z))_x= cos(5x- 4y+ z)(5- z_x)= 0. Solve for z_x.

 

[dicerandom]

HallsofIvy: I think that should be a (5+z_x) in your second expression.