Question

If y=2ln(x2+2)y=2ln⁡(x2+2), find the equation of the tangent line to the curve at x=4x=4.

Answer 1

y=21n(x2+2)y=21n(x2+2)
21n(x2+2)(y-1)=0
n=0, x=i/2-i/2, y=1