Find the equation of the normal line, to the curve y = e^x^2, when x = ln3.
Write the solution in standard form.
when x=ln3, e^x^2 = e^((ln3)^2)
and y' = 2x e^x^2
y'(ln3) = 2ln3 e^((ln3)^2)
so in point-slope form, that would be
y-e^((ln3)^2) = -1/(2ln3 e^((ln3)^2)) (x-ln3)
Or maybe you meant y=(e^x)^2 = e^(2x)
which is much easier to handle, using the same steps.