Find a formula for the nth derivative of f(x) = 4e^(3x)

To find the formula for the \( n \)-th derivative of \( f(x) = 4e^{3x} \), follow these steps:

### Step 1: Differentiate \( f(x) \) multiple times
Let's calculate the first few derivatives to identify a pattern:

1. \( f(x) = 4e^{3x} \)
   
2. First derivative (\( f'(x) \)):
   \[
   f'(x) = 4 \cdot 3e^{3x} = 12e^{3x}
   \]
   
3. Second derivative (\( f''(x) \)):
   \[
   f''(x) = 12 \cdot 3e^{3x} = 36e^{3x}
   \]
   
4. Third derivative (\( f^{(3)}(x) \)):
   \[
   f^{(3)}(x) = 36 \cdot 3e^{3x} = 108e^{3x}
   \]

### Step 2: Identify the pattern
The pattern emerges that each derivative involves multiplying by an additional factor of \( 3 \). Specifically, after differentiating \( n \) times, the factor \( 3^n \) is introduced.

Thus, the general formula for the \( n \)-th derivative is:
\[
f^{(n)}(x) = 4 \cdot 3^n \cdot e^{3x}
\]

### Final formula:
\[
f^{(n)}(x) = 4 \cdot 3^n \cdot e^{3x}
\]

Would you like more details or have any further questions?

Here are 5 related questions to explore further:
1. How would the formula change if the function was \( f(x) = e^{kx} \)?
2. What would the second derivative of \( f(x) = 5e^{4x} \) be?
3. How do you find the nth derivative of a trigonometric function like \( \sin(x) \)?
4. How does the constant \( 4 \) affect the differentiation process?
5. Can you apply this technique to \( f(x) = 7e^{-2x} \)?

**Tip:** Recognizing patterns in simple exponential functions can help you quickly derive nth-order formulas.

Discover the formula for the nth derivative of f(x) = 4e^(3x). This calculus problem involves identifying patterns in exponential functions, leading to the general formula f^{(n)}(x) = 4 \cdot 3^n \cdot e^{3x}. Ideal for students in Grades 11-12.

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