Calculus help (finding equation of a tangent line)?

Sub in the given x coordinate to find the y coordinate:

g(x) = √9/3 = 1

Recall that x = √x^2:

g(x) = (4x^-1 - 3x^-2)^1/2

Differentiate using the chain rule:

g'(x) = 1/2 * (4x^-1 - 3x^-2)^-1/2 * (-4x^-2 + 6x^-3)

g'(x) = 1/2 * (4/x - 3/x^2)^1/2 * (-4/x^2 + 6/x^3)

Sub in the given coordinates to find the slope of the tangent line:

g'(x) = 1/2 * 1 * -2/9

g'(x) = -1/9

Now that you have the slope of the tangent line, you can find its equation:

y - 1 = -1/9(x - 3)

-9y + 9 = x - 3

x + 9y - 12 = 0

g(x) = sqrt(4x - 3)/x

g'(x) = 4/(2 x sqrt(4x - 3)) - sqrt(4x - 3)/x^2 = (2/9) - 1/3 = (2 - 3)/9 = -1/9