Let’s differentiate both functions $f(x) = x \cdot 2^{3x+1}$ and $g(x) = 3^{\log(3x + 1)}$ and simplify the derivatives as much as possible.

Problem 9: Differentiate $f(x) = x \cdot 2^{3x+1}$

1. Let $f(x) = x \cdot 2^{3x+1}$.

2. Use the product rule for differentiation: $(u \cdot v)' = u'v + uv'$, where $u = x$ and $v = 2^{3x+1}$.

   • $u' = 1$.
   • For $v = 2^{3x+1}$, use the chain rule. The derivative of $2^{3x+1}$ is $2^{3x+1} \cdot \ln(2) \cdot 3$ because of the exponent’s derivative.

3. Applying the product rule: $f'(x) = (1) \cdot 2^{3x+1} + x \cdot 2^{3x+1} \cdot \ln(2) \cdot 3.$

4. Simplify: $f'(x) = 2^{3x+1} + 3x \cdot 2^{3x+1} \cdot \ln(2).$

5. Factor out $2^{3x+1}$: $f'(x) = 2^{3x+1} \left(1 + 3x \ln(2)\right).$

Problem 10: Differentiate $g(x) = 3^{\log(3x + 1)}$

1. Rewrite $g(x) = 3^{\log(3x + 1)}$.

   Notice that $3^{\log(3x + 1)} = e^{\ln(3) \cdot \log(3x + 1)}$, so we can rewrite $g(x)$ in terms of $e$ for easier differentiation if needed.

2. Differentiate $g(x)$ using the chain rule. For a function $a^{f(x)}$, the derivative is $a^{f(x)} \cdot \ln(a) \cdot f'(x)$.

3. In this case:

   • $a = 3$.
   • $f(x) = \log(3x + 1)$.

4. Differentiate $f(x) = \log(3x + 1)$: $f'(x) = \frac{3}{3x + 1}.$

5. Therefore, $g'(x) = 3^{\log(3x + 1)} \cdot \ln(3) \cdot \frac{3}{3x + 1}$.

6. Substitute back: $g'(x) = 3^{\log(3x + 1)} \cdot \frac{3 \ln(3)}{3x + 1}.$

This provides the derivatives of both functions, simplified as requested.

Would you like further clarification or details on any step? Here are some follow-up questions you might consider:

1. What are the rules for differentiating exponential functions with variable bases?
2. How does the chain rule apply when differentiating composite functions?
3. How can you verify the derivative of an exponential function involving a logarithmic expression?
4. What is the purpose of factoring out common terms in a derivative?
5. How would the differentiation process change if a different base was used in the logarithmic function?

Tip: Always check if you can simplify the expression by factoring out terms after applying the differentiation rules. This often leads to a cleaner and more interpretable answer.