find the tangent plane for the given function:
$\displaystyle \sin(xyz) = x+2y+3z, \ \ (2,-1,0)$
would I have to transfer sin(xyz) to the right then differentiate?
You can differentiate both sides partially using implicit diferentiation. Eg. Assume z is implicitly defined as a function of y and x and differentiate both sides wrt x:
$\displaystyle \cos (xyz) (yz + xy z_x) = 1 + 3z_x$.
At (2, -1, 0) substitute x = 2, y = -1 and z = 0: $\displaystyle \cos (0) (0 - 2z_x) = 1 + 3z_x \Rightarrow -2 z_x = 1 + 3 z_x \Rightarrow z_x = - \frac{1}{5}$.
Do similar to get $\displaystyle z_y$ at (2, -1, 0).
Remember that a vector normal to the tanget plane is $\displaystyle (z_x) i + (z_y) j - k$ .....