find the derivative of the functuon usinf the definition of derivative. state the domain pf the function and the domain of the dervative. g(x)= √(1+2x)

To find the derivative of the function \( g(x) = \sqrt{1 + 2x} \) using the **definition of the derivative**, we use the following formula:

\[
g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}
\]

### Step-by-step process:

1. **Substitute \( g(x) \) and \( g(x+h) \) into the definition**:

   \[
   g'(x) = \lim_{h \to 0} \frac{\sqrt{1 + 2(x+h)} - \sqrt{1 + 2x}}{h}
   \]

   Simplify \( g(x+h) \):

   \[
   g(x+h) = \sqrt{1 + 2(x+h)} = \sqrt{1 + 2x + 2h}
   \]

   So, the expression becomes:

   \[
   g'(x) = \lim_{h \to 0} \frac{\sqrt{1 + 2x + 2h} - \sqrt{1 + 2x}}{h}
   \]

2. **Rationalize the numerator** by multiplying both the numerator and denominator by the conjugate of the numerator:

   \[
   g'(x) = \lim_{h \to 0} \frac{\left( \sqrt{1 + 2x + 2h} - \sqrt{1 + 2x} \right) \cdot \left( \sqrt{1 + 2x + 2h} + \sqrt{1 + 2x} \right)}{h \cdot \left( \sqrt{1 + 2x + 2h} + \sqrt{1 + 2x} \right)}
   \]

   The numerator simplifies as a difference of squares:

   \[
   g'(x) = \lim_{h \to 0} \frac{(1 + 2x + 2h) - (1 + 2x)}{h \cdot \left( \sqrt{1 + 2x + 2h} + \sqrt{1 + 2x} \right)}
   \]

   This simplifies to:

   \[
   g'(x) = \lim_{h \to 0} \frac{2h}{h \cdot \left( \sqrt{1 + 2x + 2h} + \sqrt{1 + 2x} \right)}
   \]

3. **Cancel out \( h \)**:

   \[
   g'(x) = \lim_{h \to 0} \frac{2}{\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}}
   \]

4. **Take the limit as \( h \to 0 \)**:

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

5. **Final derivative**:

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

### Domain of \( g(x) \):
The domain of \( g(x) = \sqrt{1 + 2x} \) requires that the expression inside the square root be non-negative:

\[
1 + 2x \geq 0 \quad \Rightarrow \quad x \geq -\frac{1}{2}
\]

Thus, the domain of \( g(x) \) is \( x \in \left[ -\frac{1}{2}, \infty \right) \).

### Domain of \( g'(x) \):
Since \( g'(x) = \frac{1}{\sqrt{1 + 2x}} \), the same condition applies: \( 1 + 2x > 0 \), so \( x \geq -\frac{1}{2} \).

Thus, the domain of \( g'(x) \) is also \( x \in \left[ -\frac{1}{2}, \infty \right) \).

### Summary:
- **Derivative**: \( g'(x) = \frac{1}{\sqrt{1 + 2x}} \)
- **Domain of \( g(x) \)**: \( x \in \left[ -\frac{1}{2}, \infty \right) \)
- **Domain of \( g'(x) \)**: \( x \in \left[ -\frac{1}{2}, \infty \right) \)

Would you like a detailed explanation of any step or further details on related topics?

Here are 5 questions to expand your understanding:
1. How does rationalizing the numerator help in computing limits?
2. What are the conditions for the existence of a derivative at a point?
3. How does the chain rule apply to derivatives of composite functions like \( g(x) = \sqrt{1 + 2x} \)?
4. How would the derivative change if the function were \( g(x) = \sqrt{a + bx} \) instead?
5. What are the geometric interpretations of the derivative \( g'(x) \)?

**Tip**: Rationalizing the numerator is a common technique in calculus to simplify expressions, especially when square roots are involved.

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of the derivative. g(x) = √(1+2x).

Learn how to find the derivative of g(x) = √(1+2x) using the definition of the derivative and analyze the domain of the function and its derivative. This detailed solution walks through each step, including rationalizing the numerator and applying limits. Suitable for advanced high school calculus students.

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