Using the product rule find d/dx[g^2(x)]x=2. g(x)=4 and g’(x)=1

To find the derivative of \( g^2(x) \) at \( x = 2 \), we will apply the **product rule** to \( g(x) \cdot g(x) \).

### Step-by-step process:

1. **Identify the function**: 
   - Let \( f(x) = g(x) \cdot g(x) = g^2(x) \).
   - We are asked to find \( \frac{d}{dx} [g^2(x)] \) at \( x = 2 \), where \( g(x) = 4 \) and \( g'(x) = 1 \).

2. **Use the product rule**:  
   The product rule states:
   \[
   \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)
   \]
   In this case, \( u(x) = g(x) \) and \( v(x) = g(x) \), so:
   \[
   \frac{d}{dx} [g(x) \cdot g(x)] = g'(x) \cdot g(x) + g(x) \cdot g'(x)
   \]
   Simplifying:
   \[
   \frac{d}{dx} [g^2(x)] = 2g(x) \cdot g'(x)
   \]

3. **Substitute values at \( x = 2 \)**:
   - \( g(2) = 4 \)
   - \( g'(2) = 1 \)

   Thus, the derivative at \( x = 2 \) is:
   \[
   2 \cdot g(2) \cdot g'(2) = 2 \cdot 4 \cdot 1 = 8
   \]

### Final answer:
\[
\frac{d}{dx} [g^2(x)] \Bigg|_{x = 2} = 8
\]

Do you want further details or have any questions about the process?

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Here are 5 related questions:

1. How would the derivative change if \( g'(x) \) were negative at \( x = 2 \)?
2. Can you generalize the product rule for the derivative of \( g^n(x) \)?
3. What happens if \( g(x) \) is a constant? How would that affect the derivative?
4. How does the product rule differ from the chain rule when finding derivatives of composite functions?
5. What is the derivative of \( g^3(x) \) at \( x = 2 \) if \( g(x) = 4 \) and \( g'(x) = 1 \)?

### Tip:
The product rule is particularly useful when differentiating products of functions, but for powers of functions like \( g^n(x) \), the **chain rule** may often be more direct.

Learn how to differentiate g^2(x) at x=2 using the product rule, where g(x) = 4 and g'(x) = 1. This step-by-step solution breaks down the product rule and its application to the function. Ideal for students in Grades 11-12 or those studying calculus at the college level.

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