Calculus of a Single Variable, Chapter 2, 2.4, Section 2.4, Problem 92

You need to evaluate the second derivative of the function, hence, you need first to evaluate the first derivative, using the quotient and the chain rules, such that:
f'(x) = (1'*sqrt(x+4) - 1*(sqrt(x+4))')/((sqrt(x+4))^2)
f'(x) = (0*sqrt(x+4) - 1/(2sqrt(x+4)))/(x+4)
f'(x) = (- 1/(2sqrt(x+4)))/(x+4)
f'(x) = -1/(2(x+4)sqrt(x+4))
f'(x) = -1/(2sqrt(x+4)^3)
f'(x) = -1/(2(x+4)^(3/2))
You may evaluate the second derivative of the function using the quotiemt and the chain rules:
f''(x) = (-1'*(2(x+4)^(3/2)) +1*(2(x+4)^(3/2))')/(4(x+4)^3)
f''(x) = (0+2*(3/2)(x+4)^(3/2-1))/(4(x+4)^3)
f''(x) = (3sqrt(x+4))/(4(x+4)^3)
You need to evaluate the second derivative at the point (0,1/2), hence, you need to replace 0 for x in equation f''(x) = (3sqrt(x+4))/(4(x+4)^3) :
f''(0) = (3sqrt(0+4))/(4(0+4)^3)
f''(0) = (3sqrt(4))/(4*(4)^3) => f''(0) = (6)/(256) => f''(0) = 3/128
Hence, evaluating the second derivative at (0,1/2), yields f''(0) = 3/128.