To solve the problem $f(x) = e^x \sin(x)$ and compute the second derivative $f''(x)$, we'll proceed step by step.

Step 1: Compute the first derivative $f'(x)$

We apply the product rule, which states: $(fg)' = f' g + f g'$ Here, $f(x) = e^x$ and $g(x) = \sin(x)$.

• Derivative of $e^x$ is $e^x$.
• Derivative of $\sin(x)$ is $\cos(x)$.

Thus, using the product rule: $f'(x) = e^x \cdot \sin(x) + e^x \cdot \cos(x)$ This simplifies to: $f'(x) = e^x (\sin(x) + \cos(x))$

Step 2: Compute the second derivative $f''(x)$

Now, we need to differentiate $f'(x) = e^x (\sin(x) + \cos(x))$ again. We'll use the product rule again: $f''(x) = \left( e^x \right)' (\sin(x) + \cos(x)) + e^x \left( (\sin(x) + \cos(x))' \right)$

• The derivative of $e^x$ is still $e^x$.
• The derivative of $\sin(x) + \cos(x)$ is $\cos(x) - \sin(x)$.

Thus, the second derivative becomes: $f''(x) = e^x (\sin(x) + \cos(x)) + e^x (\cos(x) - \sin(x))$ Simplifying: $f''(x) = e^x \left[ (\sin(x) + \cos(x)) + (\cos(x) - \sin(x)) \right]$ $f''(x) = e^x (2\cos(x))$

Final Answer:

The second derivative $f''(x) = 2e^x \cos(x)$, which matches the option that was circled.

Do you want more details on this explanation, or have any questions? Here are some related questions:

1. How is the product rule applied to functions involving exponentials and trigonometric terms?
2. What are the conditions under which the chain rule should be used instead of the product rule?
3. How does $f(x) = e^x \sin(x)$ behave as $x \to \infty$?
4. What is the general pattern for higher-order derivatives of functions involving $e^x \sin(x)$?
5. Can you provide examples of using the product rule in other complex functions?

Tip: For functions like $e^x \sin(x)$, the exponential growth dominates as $x \to \infty$, but the trigonometric oscillations cause rapid changes in the behavior of the function.