Help with differentiation please?

Question

The Curve C has equation : y = 16x ^ 1/2 + 32 / x + 2
(A) Find the value of dy/dx when x = 4

(B) Find the equation of the normal to C at the point where x = 4?

I have the answers to be :
A - either 12 or 42 my guess is 12 though
B - y = -1/12x + 42 + 4/12

Answer 1

(A) dy/dx = 8x^(-1/2) - 32 / (x^2)
When x=4, dy/dx = 8*4^(-1/2) - 32 / (4^2)
= 8/2 - 32/16
= 4 - 2
= 2
(B) gradient of tangent = dy/dx =2
gradient of normal = -1/ gradient of tangent = -1/2
When x =4, y=42 (from original equation)
Equation of normal is y-42 = (-1/2)(x-4)
y = -x/2 + 44

Answer 2

a) f(x)=16x^(1/2) + 32x^(-1) + 2

f `(x) = 8x^(-1/2) - 32x^(-2)

m = f `(4) = 8 x 4^(-1/2) - 32 x 4 ^ (-2)

m = 8/4^(1/2) - 32/4² = 8/2 - 2 = 2 =dy/dx

b) f(4) = 16 x 4^(1/2) +32/4 + 2 = 42

Thus curve passes thro` (4,42)
gradient of normal at (4,42) = -1/2
equation of normal at (4,42) is given by:-
y - 42 = (-1/2)(x - 4)
y = (-1/2) x + 2 +42
y = (-1/2)x + 44
y= (-1/2)x + 44