Find \frac{dz}{dx} at (0,0,0) for sin (2x+4y+z) = 0

[Whatupdoc]

If sin (2x+4y+z) = 0 , find the first partial derivatives \frac{dz}{dx} at the point (0,0,0)

A.) \frac{dz}{dx}(0,0,0) = _________________

isnt this saying get the derivative of z, respect to x? I'm just kinda confuse since the variable 'z' is also in the problem.


well i got the derivative of that function with respect to x and got 2*cos(2x+4y+z), plugged in 0,0,0 and got 2*cos(0), which is wrong. the answer should be -2*cos(0), where did the negative sign come from?

 

[Galileo]

Judging from the question, You should regard z as a function of x (and possibly y).
So you are given that:

\sin(2x+4y+z(x,y))=0

Now use partial differentiation to find dz/dx.