Let’s solve the derivatives of the given functions one by one using the Fundamental Theorem of Calculus and its extensions.

1. $F(x) = \int_{1-3x}^2 \frac{t^3}{1 + t^3} \, dt$

The derivative of $F(x)$ involves applying the Leibniz rule: $F'(x) = \frac{d}{dx} \left( \int_{1-3x}^2 \frac{t^3}{1 + t^3} \, dt \right) = \frac{t^3}{1 + t^3} \Bigg|_{t = 2} \cdot \frac{d}{dx}(2) - \frac{t^3}{1 + t^3} \Bigg|_{t = 1-3x} \cdot \frac{d}{dx}(1-3x).$

Since the upper limit $2$ is constant, its derivative is $0$. For the lower limit: $F'(x) = -\frac{(1-3x)^3}{1 + (1-3x)^3} \cdot (-3).$

Simplify: $F'(x) = \frac{3(1-3x)^3}{1 + (1-3x)^3}.$

2. $F(x) = \int_{1}^{x} \sqrt{1 + m^3} \, dm$

Here, the limits are straightforward. Using the Fundamental Theorem of Calculus: $F'(x) = \sqrt{1 + x^3}.$

3. $F(x) = \int_{\sin x}^{\cos x} y^2 \, dy$

Using the Leibniz rule again, we differentiate: $F'(x) = y^2 \Big|_{y = \cos x} \cdot \frac{d}{dx}(\cos x) - y^2 \Big|_{y = \sin x} \cdot \frac{d}{dx}(\sin x).$

Substitute the values: $F'(x) = (\cos^2 x)(-\sin x) - (\sin^2 x)(\cos x).$

Simplify: $F'(x) = -\cos^2 x \sin x - \sin^2 x \cos x.$

Factorize: $F'(x) = -\cos x (\cos^2 x + \sin^2 x).$

Using $\cos^2 x + \sin^2 x = 1$: $F'(x) = -\cos x.$

Final Results

1. $F'(x) = \frac{3(1-3x)^3}{1 + (1-3x)^3}$
2. $F'(x) = \sqrt{1 + x^3}$
3. $F'(x) = -\cos x$

Would you like detailed explanations for each step or have further questions?

5 Related Questions:

1. How does the Leibniz rule apply to integrals with variable limits?
2. What is the Fundamental Theorem of Calculus, and how is it applied here?
3. How can you simplify derivatives with trigonometric expressions like in problem 3?
4. What are some applications of definite integrals with variable limits?
5. How does the chain rule interplay with the Fundamental Theorem in these problems