To find the derivative of the function $f(x) = \frac{1}{x + 4}$ at $x = -3$ using the difference quotient, we will apply the limit definition of the derivative:

$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

For $f(x) = \frac{1}{x + 4}$, we want to find the derivative at $x = -3$. This means $a = -3$.

Step-by-Step:

1. Write the difference quotient:

$f'(-3) = \lim_{h \to 0} \frac{f(-3 + h) - f(-3)}{h}$

2. Substitute $f(x) = \frac{1}{x + 4}$ into the equation:

$f'(-3) = \lim_{h \to 0} \frac{\frac{1}{(-3 + h) + 4} - \frac{1}{-3 + 4}}{h}$

Simplifying the expressions inside the function:

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

3. Simplify the numerator:

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

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

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

4. Simplify the expression further:

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

5. Finally, evaluate the limit as $h \to 0$:

$f'(-3) = \frac{-1}{0 + 1} = -1$

Conclusion:

The derivative of $f(x) = \frac{1}{x + 4}$ at $x = -3$ is $-1$. The limit representing this derivative is:

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

Would you like to go into more details about the difference quotient or explore other examples?

Further Exploration:

1. How do we apply the difference quotient to a quadratic function?
2. What happens if the function involves a higher-degree polynomial?
3. Can you differentiate the function $f(x) = \frac{1}{x}$ at $x = 1$ using the limit definition?
4. How does the difference quotient differ from using the power rule for differentiation?
5. What is the connection between the difference quotient and the concept of the slope of a tangent line?

Tip:

When applying the difference quotient, always try to simplify the terms in the numerator first, and look for common factors that might cancel out.