**2. Find the equation of a normal line (perpendicular to a tangent line) of**

at x = 1
What I did was solve for y which is y=2

So I have a point now (1,2)

Then I find the derivative of the function which is

y' = (1/2)(x^(-1/2)) + (-1/3)(x^(-4/3))

I then plug in 1 which is y' = 1/6 at x =1

I use the point slope equation

y - 2 = 1/6(x-1)

y = 1/6x -1/6 + 2

y = 1/6x + 11/6

Since the equation of a normal line is the negative reciprocal:

y = -1/(1/6x + 11/6)

Is this right or not I just want to confirm

Thanks in advance