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We are given the function \( f(x) = 3^x x^3 \), and we need to find its derivative.

To differentiate \( f(x) = 3^x x^3 \), we'll use the **product rule** because the function is a product of two parts: \( 3^x \) and \( x^3 \).

The product rule states that if \( f(x) = u(x)v(x) \), then:

\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\]

Here, let:
- \( u(x) = 3^x \)
- \( v(x) = x^3 \)

### Step 1: Differentiate \( u(x) = 3^x \)

The derivative of \( 3^x \) with respect to \( x \) is:

\[
u'(x) = 3^x \ln(3)
\]

### Step 2: Differentiate \( v(x) = x^3 \)

The derivative of \( x^3 \) with respect to \( x \) is:

\[
v'(x) = 3x^2
\]

### Step 3: Apply the product rule

Now apply the product rule:

\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\]

Substitute the values of \( u'(x), u(x), v'(x), \) and \( v(x) \):

\[
f'(x) = (3^x \ln(3)) x^3 + (3^x) (3x^2)
\]

### Step 4: Simplify the expression

Now simplify:

\[
f'(x) = 3^x x^3 \ln(3) + 3^x \cdot 3x^2
\]

\[
f'(x) = 3^x x^3 \ln(3) + 3 \cdot 3^x x^2
\]

So, the derivative is:

\[
f'(x) = 3^x x^2 (x \ln(3) + 3)
\]

This is the final simplified form of the derivative of \( f(x) = 3^x x^3 \).

Let me know if you need further clarifications or if you have any questions!

### Related Questions:
1. How do we derive the derivative of \( a^x \) in general?
2. Can the chain rule also be applied to this problem? Why or why not?
3. What would the second derivative of \( f(x) = 3^x x^3 \) be?
4. How does the derivative of \( 3^x \) differ from the derivative of \( x^3 \)?
5. How does the product rule compare to the quotient rule in differentiation?

### Tip:
When differentiating exponential functions like \( 3^x \), remember to multiply by \( \ln(3) \), as it's crucial for correctly applying the chain rule for exponentials!

Solve for the derivative of f(x) = 3^x * x^3

Learn how to differentiate the function f(x) = 3^x * x^3 using the product rule. This solution explains step-by-step the application of differentiation rules, including the derivative of exponential and power functions. Suitable for students in Grades 11-12.

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