Final answer:
To find the equation of the plane tangent to the surface z=ln(1+xy) at a given point, first find the partial derivatives of the function with respect to x and y. Then evaluate these partial derivatives at the given point. Finally, use the point-normal form of a plane equation to find the equation of the tangent plane.
Step-by-step explanation:
To find an equation of the plane tangent to the surface z=ln(1+xy) at a given point, we need to find the gradient (normal vector) of the surface at that point. The gradient of a function is given by the partial derivatives of the function with respect to each variable.
So, first we find the partial derivatives:
∂z/∂x = y/(1+xy)
∂z/∂y = x/(1+xy)
Next, we evaluate the partial derivatives at the given point:
∂z/∂x at (x,y) = y/(1+xy) at (x,y) = -1/(1+x)
∂z/∂y at (x,y) = x/(1+xy) at (x,y) = -x/(1+x)
Using these partial derivatives, the equation of the tangent plane can be found using the point-normal form of a plane equation:
(x-x₀)(∂z/∂x at (x₀,y₀)) + (y-y₀)(∂z/∂y at (x₀,y₀)) + (z-z₀) = 0
where (x₀,y₀,z₀) is the given point. Plugging in the values, we get the equation of the tangent plane.