Answer:
[tex]z-y+1=0[/tex]
Step-by-step explanation:
We are given that equation of surface
[tex]z=ln(x^4+y)[/tex]
Point (0,1,0)
[tex]z_x=\frac{1}{x^4+y}(4x^3)[/tex]
[tex]z_x(0,1,0)=0[/tex]
[tex]z_y=\frac{1}{x^4+y}(1)[/tex]
[tex]z_y(0,1,0)=1[/tex]
Now, the equation of tangent plane to the given surface at point (0,1,0) is given by
[tex]z-z_0=z_x(x-x_0)+z_y(y-y_0)[/tex]
Using the formula
[tex]z-0=0+1(y-1)=y-1[/tex]
[tex]z=y-1[/tex]
[tex]z-y+1=0[/tex]
