# Math Help - Tangent Plane Problem

1. ## Tangent Plane Problem

I have got stuck on finding the equation for the tangent plane for the equation:

yz = ln(x + z), at point (0, 0, 1).

What I have done is rearrange to

[ ln(x + z) ]/yz = 1, then

Fx(x, y, z) = 1/yz(x + z),

Fx(x, y, z) = 1/(xyz + yz^2)

Which I am pretty sure is right.

When it comes to sub in the x = 0, y = 0, z = 1, I get this.

1/(0 + 0), which is undefined.

What does this mean?

Cheers,

Chris

2. Differentiate with respect to y: $yz_y+z={z_y\over x+z}$; and to x: $yz_x = {1+z_x\over x+z}$. Plug in your point to get respectively $z_y=1$ and $z_x=-1$. Technically you still have to show that z is differentiable at (0,0,1), but glossing that over, you have that the tangent plane is $z-1=y-x$.