Differentiation question

A curve defined by y=(x-a) √( x- b) where a and b are constants, cuts the x -axis at A where x=b+1. Show that the gradient of the curve at A is 1

please help me with this I've been doing it all day and cant seem to arrive at the required gradient

y = (x - a)√(x - b)

x-intercept is given A(b + 1, 0)
So when x = b+1, y = 0. Set up the equation:
0 = (b + 1 - a) √(b + 1 - b)
0 = (b + 1 - a) √1
= b + 1 - a
a - b = 1

y' = √(x - b) + (x - a) (1/2) (x - b)^(-1/2)
y' = √(x - b) + (x - a)/(2√(x - b))
y' = 1/(2√(x - b)) ∙ (2x - 2b + x - a)
y' = (3x - a - 2b) / (2√(x - b))

Plug (b+1) in for x into the derivative:
y' = (3(b + 1) - a - 2b) / (2√(b + 1 - b))
y' = (3b + 3 - a - 2b) / (2√1)
y' = (b + 3 - a) / 2
y' = -(a - b - 3) / 2

Previously, we found that a - b = 1. Use this to solve for y':
y' = -(1 - 3)/2
y' = -(-2)/2
y' = 2/2
y' = 1

A is the point (b + 1, 0).
Substitute (b + 1, 0) for (x,y) and solve for a.
0 = (b + 1 - a)√(b + 1 - b)
0 = (b + 1 - a)1
0 = b + 1 - a
a = b + 1

dy/dx = 1√(x - b) + (x - a)(1/2)(x - b)^(-1/2)
dy/dx = √(x - b) + (x - a)/(2√(x - b))
dy/dx = 2(x - b)/(2√(x - b)) + (x - a)/(2√(x - b))
dy/dx = (2x - 2b + x - a)/(2√(x - b))
dy/dx(at (b + 1, 0)) = (2(b + 1) - 2b + (b + 1) - a)/(2√(b + 1 - b)) =
(2b + 2 - 2b + b + 1 - a)/(2√(b + 1 - b)) =
(3 + b - a)/(2√(b + 1 - b) =...substitute b + 1 for a next
(3 + b - (b + 1))/2 =
(3 - 1)/2 =
1