Let's go through the problem and find the derivative of each function step by step:

1. $f(x) = x^3 - 2x$

   To differentiate this: $f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x)$ Using the power rule: $f'(x) = 3x^2 - 2$

2. $f(x) = \sqrt{x}$

   Recall that $\sqrt{x} = x^{1/2}$, so we apply the power rule: $f'(x) = \frac{d}{dx} \left( x^{1/2} \right) = \frac{1}{2}x^{-1/2}$ $f'(x) = \frac{1}{2\sqrt{x}}$

3. $f(x) = \frac{2 - x}{1 + x}$

   Here, we apply the quotient rule, which states that for a function $\frac{u(x)}{v(x)}$, the derivative is: $f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$ Let $u(x) = 2 - x$ and $v(x) = 1 + x$.

   First, compute the derivatives of $u(x)$ and $v(x)$: $u'(x) = -1, \quad v'(x) = 1$ Now apply the quotient rule: $f'(x) = \frac{(1 + x)(-1) - (2 - x)(1)}{(1 + x)^2}$ Simplify: $f'(x) = \frac{-(1 + x) - (2 - x)}{(1 + x)^2} = \frac{-1 - x - 2 + x}{(1 + x)^2} = \frac{-3}{(1 + x)^2}$

4. $f(x) = \sin(x)$

   The derivative of $\sin(x)$ is: $f'(x) = \cos(x)$

5. $f(x) = \cos(x)$

   The derivative of $\cos(x)$ is: $f'(x) = -\sin(x)$

Summary of derivatives:

1. $f'(x) = 3x^2 - 2$
2. $f'(x) = \frac{1}{2\sqrt{x}}$
3. $f'(x) = \frac{-3}{(1 + x)^2}$
4. $f'(x) = \cos(x)$
5. $f'(x) = -\sin(x)$

Would you like further details on any step?

Here are 5 related questions to expand on this:

1. What is the power rule, and how is it applied in derivatives?
2. How is the quotient rule used to differentiate more complex functions?
3. What are the key rules for differentiating trigonometric functions like $\sin(x)$ and $\cos(x)$?
4. Can you derive a general formula for differentiating $\frac{a + bx}{c + dx}$?
5. How do the product rule and chain rule work in combination for more complex functions?

Tip: Always simplify your function, if possible, before applying derivative rules for more efficient computation!