\( \frac { d } { d x } \int _ { x } ^ { 4 } \sqrt { t ^ { 3 } + 2 } d t = \square \)

The derivative of the integral with respect to \( x \) cannot be calculated directly. However, we can simplify the expression by using the Fundamental Theorem of Calculus, which states that the derivative of an integral is equal to the integrand evaluated at the upper limit of integration. Therefore, the derivative of the integral with respect to \( x \) is: \[ \frac{d}{dx} \int_{x}^{4} \sqrt{t^{3}+2} dt = \sqrt{4^{3}+2} \] \[ \frac{d}{dx} \int_{x}^{4} \sqrt{t^{3}+2} dt = \sqrt{64+2} \] \[ \frac{d}{dx} \int_{x}^{4} \sqrt{t^{3}+2} dt = \sqrt{66} \] Therefore, the derivative of the integral with respect to \( x \) is \( \sqrt{66} \).