1. **State the problem:** Find the derivative with respect to $x$ of the integral $$\frac{d}{dx} \int_0^{\sqrt{x}} \cos t \, dt.$$ 2. **Recall the Fundamental Theorem of Calculus and Chain Rule:** If $$F(x) = \int_a^{g(x)} f(t) \, dt,$$ then $$F'(x) = f(g(x)) \cdot g'(x).$$ 3. **Apply the formula:** Here, $$f(t) = \cos t,$$ and the upper limit is $$g(x) = \sqrt{x} = x^{1/2}.$$ 4. **Compute the derivative of the upper limit:** $$g'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}.$$ 5. **Evaluate the derivative:** $$\frac{d}{dx} \int_0^{\sqrt{x}} \cos t \, dt = \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\cos(\sqrt{x})}{2\sqrt{x}}.$$ 6. **Final answer:** $$\boxed{\frac{d}{dx} \int_0^{\sqrt{x}} \cos t \, dt = \frac{\cos(\sqrt{x})}{2\sqrt{x}}}.$$