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It looks like you're working on a calculus worksheet involving derivatives by first principles. I'll guide you through solving these problems.

### First principles formula:
The derivative \( f'(x) \) by first principles is given by the limit:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

Now let's approach both parts:

### Part (a)
The function provided is:

\[
f(x) = \frac{1}{x - 3}
\]

Using the first principles, we need to evaluate:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

Substitute \( f(x) \) and \( f(x+h) \):

\[
f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)-3} - \frac{1}{x-3}}{h}
\]

Simplify the numerator:

\[
= \lim_{h \to 0} \frac{\frac{1}{x+h-3} - \frac{1}{x-3}}{h} = \lim_{h \to 0} \frac{(x-3)-(x+h-3)}{(x+h-3)(x-3)} \cdot \frac{1}{h}
\]

\[
= \lim_{h \to 0} \frac{-h}{(x+h-3)(x-3) \cdot h}
\]

Cancel the \( h \) in the numerator and denominator:

\[
= \lim_{h \to 0} \frac{-1}{(x+h-3)(x-3)}
\]

As \( h \to 0 \), the expression simplifies to:

\[
f'(x) = \frac{-1}{(x-3)^2}
\]

### Part (b)
The function provided is:

\[
f(x) = \frac{1}{x + 2}
\]

Again, using the first principles:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

Substitute \( f(x) \) and \( f(x+h) \):

\[
f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h+2} - \frac{1}{x+2}}{h}
\]

Simplify the numerator:

\[
= \lim_{h \to 0} \frac{(x+2)-(x+h+2)}{(x+h+2)(x+2)} \cdot \frac{1}{h}
\]

\[
= \lim_{h \to 0} \frac{-h}{(x+h+2)(x+2) \cdot h}
\]

Cancel the \( h \):

\[
= \lim_{h \to 0} \frac{-1}{(x+h+2)(x+2)}
\]

As \( h \to 0 \), this simplifies to:

\[
f'(x) = \frac{-1}{(x+2)^2}
\]

### Final answers:
- (a) \( f'(x) = \frac{-1}{(x-3)^2} \)
- (b) \( f'(x) = \frac{-1}{(x+2)^2} \)

Would you like more detailed steps for either part, or explanations on a specific step?

Here are 5 related questions for practice:
1. How does the first principles method differ from using derivative rules?
2. Can you find the derivative of \( f(x) = \frac{1}{x+1} \) using first principles?
3. What is the derivative of \( f(x) = \frac{1}{x^2 + 1} \) by first principles?
4. How would you approach finding the derivative of a quadratic function by first principles?
5. Can you explain the concept of limits as used in the first principles definition of the derivative?

**Tip:** When working with limits in first principles, always aim to simplify the algebra before taking the limit to zero for cleaner results.

Find f'(x) by first principles for the functions: f(x) = 1/(x - 3) and f(x) = 1/(x + 2).

Learn how to find the derivative of functions like 1/(x - 3) and 1/(x + 2) using first principles. This step-by-step guide focuses on the fundamental calculus concepts of limits and derivatives, perfect for Grade 12 or first-year college students.

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