derivative of (x^3/6)+(1/2x)?

i got (x^2/2)-(1/2x^2) is that correct?
and then wut is the (F'(x))2?...meaning the derivative squred?

if you meant (1/2)x^2 - 1/(2x^2) you are correct

I don't think they are looking for the square of the derivative, but rather the second derivative.
just differentiate again what you had in your first derivative.

no i am looking for the derivattive squared becasue the accual problem is to find the surface area and to find it i need (f(x)^2)

I apologize for the misunderstanding. You are correct in that to find the surface area, you need to compute the square of the derivative, also known as the second derivative, denoted as (f'(x))^2.

To find the second derivative of the function f(x) = (x^3/6) + (1/2x), we can proceed as follows:

Step 1: Find the first derivative, f'(x), using the rules of differentiation:
f'(x) = (1/6) * 3x^2 - (1/2) * (1/x^2)
= (1/2)x^2 - (1/2) / x^2

Step 2: Square the first derivative:
(f'(x))^2 = [(1/2)x^2 - (1/2) / x^2]^2
= (1/4)x^4 - (1/2x^2) + (1/4) / x^4
= (1/4)x^4 - (1/2x^2) + (1/4) * (1 / x^4)
= (1/4)x^4 - (1/2x^2) + 1 / (4x^4)

So, the expression for the square of the derivative is (f'(x))^2 = (1/4)x^4 - (1/2x^2) + 1 / (4x^4).

Please note that the second derivative refers to the derivative of the first derivative. The derivative squared usually implies squaring the original function rather than the derivative.