To solve Exercise 26, we need to find the derivative of the function $g(x) = 2 + 2\sqrt{x}$ using the alternative formula for derivatives:

$g'(x) = \lim_{z \to x} \frac{g(z) - g(x)}{z - x}$

Step-by-Step Solution

1. Substitute $g(x)$ and $g(z)$ in the formula:

   $g(x) = 2 + 2\sqrt{x} \quad \text{and} \quad g(z) = 2 + 2\sqrt{z}$

   Therefore,

   $g(z) - g(x) = (2 + 2\sqrt{z}) - (2 + 2\sqrt{x}) = 2\sqrt{z} - 2\sqrt{x}$

2. Rewrite the limit expression:

   Substitute into the limit expression for $g'(x)$:

   $g'(x) = \lim_{z \to x} \frac{2\sqrt{z} - 2\sqrt{x}}{z - x}$

3. Factor out 2:

   $g'(x) = \lim_{z \to x} \frac{2(\sqrt{z} - \sqrt{x})}{z - x}$

4. Use the difference of squares to simplify:

   To simplify $\sqrt{z} - \sqrt{x}$, multiply the numerator and the denominator by the conjugate $\sqrt{z} + \sqrt{x}$:

   $g'(x) = \lim_{z \to x} \frac{2(\sqrt{z} - \sqrt{x})(\sqrt{z} + \sqrt{x})}{(z - x)(\sqrt{z} + \sqrt{x})}$

   Using the difference of squares in the numerator:

   $= \lim_{z \to x} \frac{2(z - x)}{(z - x)(\sqrt{z} + \sqrt{x})}$

5. Cancel $z - x$ from the numerator and denominator:

   $g'(x) = \lim_{z \to x} \frac{2}{\sqrt{z} + \sqrt{x}}$

6. Take the limit as $z \to x$:

   Substitute $z = x$:

   $g'(x) = \frac{2}{\sqrt{x} + \sqrt{x}} = \frac{2}{2\sqrt{x}} = \frac{1}{\sqrt{x}}$

Final Answer

$g'(x) = \frac{1}{\sqrt{x}}$

Would you like further details on this solution or have any additional questions?

Here are 5 related questions that expand on this topic:

1. How would the derivative formula change if the function included a different square root term, like $g(x) = 3 + \sqrt{3x}$?
2. How can we apply this method to find the derivative of a function like $h(x) = x^{1/3}$?
3. What happens if the function has a cube root, such as $g(x) = 2 + 2\sqrt[3]{x}$?
4. Can we apply the alternative formula for derivatives to polynomial functions, like $f(x) = x^2 + x$?
5. How does the alternative formula for derivatives compare to the standard limit definition of a derivative?

Tip: Multiplying by the conjugate is a helpful technique when dealing with square roots in derivative problems, as it helps simplify complex expressions.