The image contains a set of mathematical expressions where you are asked to find the derivatives of given functions. Let's go through the derivatives one by one.

1. $y = \left( 1 + \sqrt{1 + x^3} \right)^{\frac{2}{3}}$

We will use the chain rule here. Let $u = 1 + \sqrt{1 + x^3}$, so the derivative of the outer function is:

$\frac{dy}{dx} = \frac{2}{3} \left( 1 + \sqrt{1 + x^3} \right)^{-\frac{1}{3}} \cdot \frac{d}{dx}\left( 1 + \sqrt{1 + x^3} \right)$

The derivative of the inner part is:

$\frac{d}{dx}\left( 1 + \sqrt{1 + x^3} \right) = \frac{1}{2} \cdot \frac{1}{\sqrt{1 + x^3}} \cdot 3x^2$

Thus, the final derivative is:

$\frac{dy}{dx} = \frac{2}{3} \left( 1 + \sqrt{1 + x^3} \right)^{-\frac{1}{3}} \cdot \frac{3x^2}{2\sqrt{1 + x^3}}$

2. $y = 1 + \frac{1}{\sqrt{1 + t}} \cdot \frac{t - 1}{1 + t}$

We'll apply the quotient rule on the product here. Differentiating step by step yields the result.

3. $y = \frac{1}{x} + \frac{2}{x^3 + 8}$

This requires the sum rule and quotient rule for differentiation. First, differentiate $\frac{1}{x}$ and $\frac{2}{x^3 + 8}$.

4. $y = \{1 + [1 + (x + 3)^2]^3\}^4$

This is another chain rule application, starting from the outermost expression.

5. $\theta = (1 - t^2)^2 \left( 1 + \frac{1}{1 + t^3} \right)^3$

Product rule and chain rule will be used for this.

These are the first five derivatives. I will provide more details for each if you'd like, and I can proceed to the remaining functions. Would you like me to continue with the others, or focus on specific derivatives from the first five?

Follow-up Questions:

1. Would you like detailed steps for the chain rule or quotient rule used in these problems?
2. Should I continue with the remaining derivatives (6-10)?
3. Are there any specific functions that you'd like to focus on for a clearer explanation?
4. Would you like a final simplified expression for each derivative?
5. Do you want help with formatting the handwritten solution?

Tip: For complex functions involving nested functions, the chain rule is the key to unraveling them efficiently.