Differentiate with respect to y: ; and to x: . Plug in your point to get respectively and . 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 .
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