For any n, if f(x) = x^n, f'(x) = nx^(n-1)
1) f(x) = 1/2 * x^-1
2) f(x) = 1/6 * x^-2
The constants out front don't change. When you apply the differentiation rule, you simply need to multiply by the constant in front.
For example, the first question:
f(x) = 1/2 * x^-1. In this case, n = -1 --> f'(x) = 1/2 * (-1) * x^-2 = -1/2 * x^-2. f"(x) = -1/2 * -2 * x^-3 = x^-3