https://file.aiquickdraw.com/mathbot/file/1730022196054kdqbsgek.jpg

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.

Solve Exercise 26: Find the derivative of the function g(x) = 2 + 2√x using the alternative formula for derivatives.

Learn how to find the derivative of g(x) = 2 + 2√x using the alternative derivative formula with a step-by-step solution. This problem covers key calculus concepts like limits, square roots, and simplifying expressions using the difference of squares. Ideal for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation