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Let's go through the derivatives of the given functions step by step.

### **23.** \( f(x) = \frac{1}{x + 2} \)

This is a quotient, and we can use the **quotient rule** or rewrite the function as \( (x + 2)^{-1} \).

Using the **chain rule**:
\[
f'(x) = -1 \cdot (x + 2)^{-2} \cdot \frac{d}{dx}(x + 2) = -\frac{1}{(x + 2)^2}
\]

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### **24.** \( f(x) = x^2 - 3x + 4 \)

This is a polynomial function. We apply the **power rule**:
\[
f'(x) = 2x - 3
\]

---

### **25.** \( g(x) = \frac{x}{x - 1} \)

We apply the **quotient rule** here:
\[
\left( \frac{u}{v} \right)' = \frac{v u' - u v'}{v^2}
\]
where \( u = x \) and \( v = x - 1 \). Now:
- \( u' = 1 \)
- \( v' = 1 \)

Thus:
\[
g'(x) = \frac{(x - 1)(1) - (x)(1)}{(x - 1)^2} = \frac{x - 1 - x}{(x - 1)^2} = \frac{-1}{(x - 1)^2}
\]

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### **26.** \( g(x) = 1 + \sqrt{x} \)

Rewrite \( \sqrt{x} \) as \( x^{1/2} \) and apply the **power rule**:
\[
g'(x) = 0 + \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}
\]

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Do you need further explanations on any of these solutions?

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### Related questions:
1. How does the product rule work, and how is it different from the quotient rule?
2. What is the chain rule, and when should you apply it?
3. How do you find higher-order derivatives of a function?
4. What happens to the derivative when the function contains logarithmic terms?
5. How do you apply implicit differentiation to solve equations involving multiple variables?

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**Tip:** Always simplify functions before differentiating them. It can often save you time and reduce errors.

Find the derivatives of the functions: f(x) = 1 / (x + 2), f(x) = x^2 - 3x + 4, g(x) = x / (x - 1), g(x) = 1 + √x.

Explore the solutions for differentiating functions like f(x) = 1 / (x + 2), f(x) = x^2 - 3x + 4, g(x) = x / (x - 1), and g(x) = 1 + √x. This includes applications of the Quotient Rule, Power Rule, and Chain Rule, ideal for advanced calculus students.

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