Answer is: option2
1.350Solution:
Using the chain rule, differentiate \( f(x) \): \[ f'(x) = 5 \left( 1 + \frac{x}{20} \right)^4 \cdot \frac{1}{20} \] \[ f'(x) = \frac{5}{20} \left( 1 + \frac{x}{20} \right)^4 = \frac{1}{4} \left( 1 + \frac{x}{20} \right)^4 \] Differentiate again: \[ f''(x) = \frac{1}{4} \cdot 4 \left( 1 + \frac{x}{20} \right)^3 \cdot \frac{1}{20} \] \[ f''(x) = \frac{1}{20} \left( 1 + \frac{x}{20} \right)^3 \] Substituting \( x = 40 \): \[ f''(40) = \frac{1}{20} \left( 1 + \frac{40}{20} \right)^3 \] \[ = \frac{1}{20} (1 + 2)^3 \] \[ = \frac{1}{20} (3)^3 \] \[ = \frac{27}{20} \] \[ = 1.35 \] Final Answer: 1.350