Find an equation for the tangent line to the curve at the given point. y=x^2+(lnx)^2, (1,1)

Question

please answer 

Answer 1

Equation of the tangent: y=kx+b, where k=f'(x₀).

y'=2(x+lnx/x), 

y'(1)=2=k

y=2x+b

1=2+b, b=-1

y=2x-1