To find the instantaneous rate of change of the function $f(x) = \sqrt{x + 5}$ at $x = 2$, we can use the limit definition of the derivative. The formula for the derivative using the limit definition is:

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

Let's apply this to $f(x) = \sqrt{x + 5}$ at $x = 2$.

1. First, substitute $x = 2$ into the limit definition: $f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}$

2. Now calculate $f(2)$ and $f(2 + h)$:

   • $f(2) = \sqrt{2 + 5} = \sqrt{7}$
   • $f(2 + h) = \sqrt{(2 + h) + 5} = \sqrt{h + 7}$

3. Substitute these into the limit definition: $f'(2) = \lim_{h \to 0} \frac{\sqrt{h + 7} - \sqrt{7}}{h}$

4. To simplify, multiply the numerator and denominator by the conjugate of the numerator: $f'(2) = \lim_{h \to 0} \frac{(\sqrt{h + 7} - \sqrt{7}) (\sqrt{h + 7} + \sqrt{7})}{h(\sqrt{h + 7} + \sqrt{7})}$

   Using the difference of squares: $f'(2) = \lim_{h \to 0} \frac{(h + 7) - 7}{h(\sqrt{h + 7} + \sqrt{7})} = \lim_{h \to 0} \frac{h}{h(\sqrt{h + 7} + \sqrt{7})}$

5. Simplify: $f'(2) = \lim_{h \to 0} \frac{1}{\sqrt{h + 7} + \sqrt{7}}$

6. Now, as $h \to 0$, $\sqrt{h + 7} \to \sqrt{7}$. So: $f'(2) = \frac{1}{\sqrt{7} + \sqrt{7}} = \frac{1}{2\sqrt{7}}$

Thus, the instantaneous rate of change of $f(x) = \sqrt{x + 5}$ at $x = 2$ is:

$f'(2) = \frac{1}{2\sqrt{7}}$

Would you like further details on this, or do you have any questions?

Here are some related questions:

1. How do you interpret the meaning of the instantaneous rate of change?
2. Can the conjugate method be used to simplify limits of other types of functions?
3. How would you find the rate of change at a different value of $x$?
4. What happens to the limit if you use a different approach like differentiation rules instead?
5. How does this compare to the average rate of change over an interval?

Tip: The conjugate method is a powerful technique for simplifying limits involving square roots.