Question

Find an equation of the tangent line to the curve at the point (1, 1).
y = ln (xe^x^2)

Answer 1

The equation of the tangent line is
y-yo = f'(xo) (x-xo)
f'(y) = (e^{x^2} + 2x*xe^{x^2})/(xe^{x^2}) = (1+ 2x²)/x
f(1) = 3
y -1 = 3(x-1)
y = 3x -2.