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.