Tangent plane, directional derivatives

[Laura1321412]

Homework Statement

find the equation on the tangent plane of yz=ln(x+z) at point (0, 0, 1 )

Homework Equations

Tangent plane equation...

The Attempt at a Solution

I wasn't sure how to determine the partials on this equation. My attempt was to rearange as ln(x+z)-yz=0 so Fx = 1/(x+z) Fy= -1 Fz = 1/(x+z) at the point (001) Fx= 1 Fy= -1 Fz= 1

into the plane equations and i get

x-y+z-1=0

... I am not sure if the answer is right, i think the issue is when i rearanged the equation, how can i determine the partials on this correctly?

Thank you!

 

[Laura1321412]

Hello? :(

 

[Dick]

You are finding the gradient vector to the surface F(x,y,z)=ln(x+z)-yz=0. Some of your components aren't quite correct for a general x,y,z but they are correct for (x,y,z)=(0,0,1). So yes, the normal vector is (1,-1,1) and the plane has to pass through (0,0,1) so I think your answer is correct. Can you correct some of the 'partials'?