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The problem asks to find \( g'(x) \) using the definition of the derivative for the function \( g(x) = \sqrt{15x} \). Then, it asks to find the values of the derivative at specific points: \( g'(-3) \), \( g'(0) \), and \( g'(4) \), where the derivative exists.

### Step 1: Definition of the derivative
The derivative of a function \( g(x) \) at any point \( x \) is defined as:
\[
g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}
\]
For \( g(x) = \sqrt{15x} \), let's substitute this into the formula.

### Step 2: Applying the definition of the derivative
Substitute \( g(x) = \sqrt{15x} \) into the definition:
\[
g'(x) = \lim_{h \to 0} \frac{\sqrt{15(x+h)} - \sqrt{15x}}{h}
\]
Now, simplify the expression using the technique of multiplying by the conjugate:
\[
g'(x) = \lim_{h \to 0} \frac{\left( \sqrt{15(x+h)} - \sqrt{15x} \right)\left( \sqrt{15(x+h)} + \sqrt{15x} \right)}{h \cdot \left( \sqrt{15(x+h)} + \sqrt{15x} \right)}
\]
This simplifies to:
\[
g'(x) = \lim_{h \to 0} \frac{15(x+h) - 15x}{h \cdot \left( \sqrt{15(x+h)} + \sqrt{15x} \right)}
\]
\[
g'(x) = \lim_{h \to 0} \frac{15h}{h \cdot \left( \sqrt{15(x+h)} + \sqrt{15x} \right)}
\]
Cancel out the \( h \) terms:
\[
g'(x) = \lim_{h \to 0} \frac{15}{\sqrt{15(x+h)} + \sqrt{15x}}
\]
As \( h \to 0 \), the expression simplifies to:
\[
g'(x) = \frac{15}{2\sqrt{15x}}
\]
Thus, the derivative is:
\[
g'(x) = \frac{1}{2\sqrt{x}}
\]

### Step 3: Evaluate \( g'(x) \) at specific points
Now, we compute \( g'(-3) \), \( g'(0) \), and \( g'(4) \) where the derivative exists.

1. **At \( x = -3 \)**:
   The function \( g(x) = \sqrt{15x} \) is undefined for negative values of \( x \), so \( g'(-3) \) does not exist.

2. **At \( x = 0 \)**:
   Substituting \( x = 0 \) into \( g'(x) = \frac{1}{2\sqrt{x}} \):
   \[
   g'(0) = \frac{1}{2\sqrt{0}} = \text{undefined}
   \]
   So, \( g'(0) \) does not exist because the square root of zero in the denominator makes the derivative undefined.

3. **At \( x = 4 \)**:
   Substituting \( x = 4 \) into \( g'(x) = \frac{1}{2\sqrt{x}} \):
   \[
   g'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{2 \times 2} = \frac{1}{4}
   \]
   So, \( g'(4) = \frac{1}{4} \).

### Conclusion:
- \( g'(x) = \frac{1}{2\sqrt{x}} \)
- \( g'(-3) \) does not exist.
- \( g'(0) \) does not exist.
- \( g'(4) = \frac{1}{4} \).

Would you like further details or have any questions?

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Here are five related questions:
1. What are the conditions for the existence of a derivative at a point?
2. How does multiplying by the conjugate help simplify derivative calculations?
3. What is the domain of the function \( g(x) = \sqrt{15x} \)?
4. Why is the derivative undefined at \( x = 0 \)?
5. How would the derivative change for a different square root function?

**Tip:** When dealing with square root functions, always check the domain, as the square root is only defined for non-negative inputs.

Using the definition of the derivative: Find g'(x). Then find g'(-3), g'(0), and g'(4) when the derivative exists. g(x) = √15x

Learn how to find the derivative of g(x) = √15x using the definition of the derivative. This detailed step-by-step solution includes the calculation of g'(-3), g'(0), and g'(4), highlighting key calculus concepts and theorems. Suitable for students in Grades 11-12.

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