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Let's go through each of the problems one by one and find the derivatives \( f'(x) \), then evaluate at the given points.

### 1. For \( f(x) = x^2 - x \):
- The derivative is:
\[
f'(x) = \frac{d}{dx}(x^2 - x) = 2x - 1
\]
- Evaluate at \( x = 5 \):
\[
f'(5) = 2(5) - 1 = 10 - 1 = 9
\]
So, \( f'(5) = 9 \).

### 2. For \( f(x) = x^3 \):
- The derivative is:
\[
f'(x) = \frac{d}{dx}(x^3) = 3x^2
\]
- Evaluate at \( x = -2 \):
\[
f'(-2) = 3(-2)^2 = 3(4) = 12
\]
So, \( f'(-2) = 12 \).

### 3. For \( f(x) = \frac{1}{x+1} \):
- Using the chain rule and the power rule:
\[
f'(x) = \frac{d}{dx}\left(\frac{1}{x+1}\right) = -\frac{1}{(x+1)^2}
\]
- Evaluate at \( x = 3 \):
\[
f'(3) = -\frac{1}{(3+1)^2} = -\frac{1}{4^2} = -\frac{1}{16}
\]
So, \( f'(3) = -\frac{1}{16} \).

### Final Answers:
1. \( f'(5) = 9 \)
2. \( f'(-2) = 12 \)
3. \( f'(3) = -\frac{1}{16} \)

Would you like further explanation on any of these steps?

Here are five related questions:
1. How does the power rule apply when differentiating polynomials?
2. Can you explain the chain rule in more detail, especially in the third problem?
3. What is the geometric interpretation of a derivative?
4. How would the derivative change for functions involving more complex fractions?
5. Can the product or quotient rule be applied to solve these problems?

**Tip:** When dealing with rational functions, always check if the chain rule applies for simpler differentiation.

Find the derivative f'(x) for the given functions and evaluate at specific points: 1. f'(5) for f(x) = x^2 - x, 2. f'(-2) for f(x) = x^3, 3. f'(3) for f(x) = 1/(x+1).

This problem involves finding the derivatives of polynomial and rational functions, applying rules such as the power rule and chain rule. Step-by-step solutions for f'(5), f'(-2), and f'(3) are provided, along with explanations of key differentiation techniques. Suitable for Grades 11-12 or introductory calculus.

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