d. Use the Fundamental theorem of Calculus to compute the derivative of \( F(x)= \) \( \left(\int_{0}^{x} \cos (x) d t\right)^{3} \).

To compute the derivative of the function \[ F(x) = \left(\int_{0}^{x} \cos(t) \, dt\right)^{3}, \] we will use the Fundamental Theorem of Calculus and the chain rule. First, we need to find the integral \[ G(x) = \int_{0}^{x} \cos(t) \, dt. \] According to the Fundamental Theorem of Calculus, the derivative of \( G(x) \) is \[ G'(x) = \cos(x). \] Now, we can express \( F(x) \) in terms of \( G(x) \): \[ F(x) = (G(x))^3. \] Next, we will differentiate \( F(x) \) using the chain rule: \[ F'(x) = 3(G(x))^2 \cdot G'(x). \] Substituting \( G'(x) = \cos(x) \) into the equation, we have: \[ F'(x) = 3(G(x))^2 \cdot \cos(x). \] Now we need to substitute back \( G(x) \): \[ F'(x) = 3\left(\int_{0}^{x} \cos(t) \, dt\right)^{2} \cdot \cos(x). \] Thus, the derivative of \( F(x) \) is \[ F'(x) = 3\left(\int_{0}^{x} \cos(t) \, dt\right)^{2} \cdot \cos(x). \]