Given
f ( x )

To find the derivative of the function $f(x) = \sqrt{x} - 6$ at $x = 7$ using the definition of a derivative, we need to apply the limit definition of a derivative:

$f'(7) = \lim_{h \to 0} \frac{f(7+h) - f(7)}{h}$

Let's break this down step by step:

1. Evaluate $f(7)$: $f(7) = \sqrt{7} - 6$

2. Evaluate $f(7 + h)$: $f(7 + h) = \sqrt{7 + h} - 6$

3. Set up the difference quotient: $\frac{f(7 + h) - f(7)}{h} = \frac{\left( \sqrt{7 + h} - 6 \right) - \left( \sqrt{7} - 6 \right)}{h}$ Simplify the numerator: $= \frac{\sqrt{7 + h} - \sqrt{7}}{h}$

4. Multiply numerator and denominator by the conjugate: To simplify this expression, we multiply both the numerator and denominator by the conjugate of the numerator: $\frac{\sqrt{7 + h} - \sqrt{7}}{h} \cdot \frac{\sqrt{7 + h} + \sqrt{7}}{\sqrt{7 + h} + \sqrt{7}} = \frac{\left( \sqrt{7 + h} - \sqrt{7} \right) \left( \sqrt{7 + h} + \sqrt{7} \right)}{h \left( \sqrt{7 + h} + \sqrt{7} \right)}$ Simplify the numerator using the difference of squares: $= \frac{(7 + h) - 7}{h \left( \sqrt{7 + h} + \sqrt{7} \right)} = \frac{h}{h \left( \sqrt{7 + h} + \sqrt{7} \right)}$

5. Simplify and take the limit: Cancel out $h$ in the numerator and denominator: $= \frac{1}{\sqrt{7 + h} + \sqrt{7}}$ Now, take the limit as $h \to 0$: $\lim_{h \to 0} \frac{1}{\sqrt{7 + h} + \sqrt{7}} = \frac{1}{\sqrt{7} + \sqrt{7}} = \frac{1}{2\sqrt{7}}$

Therefore, the derivative of $f(x) = \sqrt{x} - 6$ at $x = 7$ is: $f'(7) = \frac{1}{2\sqrt{7}}$

Would you like to go deeper into any step or have any questions?

---

Here are 5 related questions for further practice:

1. What is the derivative of $f(x) = \sqrt{x} - 6$ at $x = 9$?
2. How would you find the derivative of $f(x) = \sqrt{x}$ at a general point $x = a$?
3. Can you apply the same method to find the derivative of $f(x) = x^2 - 5x$?
4. How can you use the chain rule to differentiate $f(x) = \sqrt{2x + 1}$?
5. What are the advantages of using the definition of a derivative over shortcut rules?

Tip: Always try to simplify the expressions before plugging in the limit for $h \to 0$ when using the definition of a derivative.