Let's solve both questions by differentiating each function with respect to $x$.

Question 12

The function given is: $y = e^x \cdot 3^x$ To differentiate $y$ with respect to $x$, we use the product rule: $\frac{d}{dx}(u \cdot v) = u'v + uv'$ where $u = e^x$ and $v = 3^x$.

1. Differentiate $u = e^x$: $u' = e^x$

2. Differentiate $v = 3^x$: Since $3^x = e^{x \ln 3}$, we can differentiate it as: $v' = 3^x \ln 3$

Now, apply the product rule: $y' = e^x \cdot 3^x \ln 3 + e^x \cdot 3^x$ Factor out $e^x \cdot 3^x$: $y' = e^x \cdot 3^x (\ln 3 + 1)$

Question 13

The function given is: $y = e^x \cdot \ln(1 + x^2)$ Again, we use the product rule where $u = e^x$ and $v = \ln(1 + x^2)$.

1. Differentiate $u = e^x$: $u' = e^x$

2. Differentiate $v = \ln(1 + x^2)$: Using the chain rule: $v' = \frac{1}{1 + x^2} \cdot 2x = \frac{2x}{1 + x^2}$

Now, apply the product rule: $y' = e^x \cdot \ln(1 + x^2) + e^x \cdot \frac{2x}{1 + x^2}$ Factor out $e^x$: $y' = e^x \left( \ln(1 + x^2) + \frac{2x}{1 + x^2} \right)$

Summary of Answers

1. For Question 12: $y' = e^x \cdot 3^x (\ln 3 + 1)$

2. For Question 13: $y' = e^x \left( \ln(1 + x^2) + \frac{2x}{1 + x^2} \right)$

Would you like further details on these steps?

Here are some additional related questions to consider:

1. How would you differentiate a function with a product of three terms?
2. What is the derivative of $3^x$ in terms of its base and exponent?
3. How does the chain rule apply to logarithmic functions?
4. Can we simplify the expression for $y'$ in Question 13 further?
5. How does the natural logarithm's derivative affect complex functions?

Tip: For exponential and logarithmic functions, always remember to apply the chain rule when there’s a composite function inside.