The given expression to differentiate is:

$\left( \cos(2x) + 1 \right)^2 \cdot e^{4x}$

To differentiate this, we need to apply the product rule. The product rule states that for two functions $u(x)$ and $v(x)$, the derivative of their product is given by:

$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

Here, we can set:

• $u(x) = \left( \cos(2x) + 1 \right)^2$
• $v(x) = e^{4x}$

Step 1: Differentiate $u(x) = \left( \cos(2x) + 1 \right)^2$

Using the chain rule: $\frac{d}{dx} \left( \cos(2x) + 1 \right)^2 = 2 \left( \cos(2x) + 1 \right) \cdot \frac{d}{dx} \left( \cos(2x) + 1 \right)$ $= 2 \left( \cos(2x) + 1 \right) \cdot (-\sin(2x) \cdot 2)$ $= -4 \left( \cos(2x) + 1 \right) \sin(2x)$

Step 2: Differentiate $v(x) = e^{4x}$

$\frac{d}{dx} e^{4x} = 4e^{4x}$

Step 3: Apply the product rule

Now apply the product rule: $\frac{d}{dx} \left[ \left( \cos(2x) + 1 \right)^2 \cdot e^{4x} \right] = \left( -4 \left( \cos(2x) + 1 \right) \sin(2x) \right) \cdot e^{4x} + \left( \cos(2x) + 1 \right)^2 \cdot 4e^{4x}$

Simplifying, the final expression is:

$= -4 \left( \cos(2x) + 1 \right) \sin(2x) e^{4x} + 4 \left( \cos(2x) + 1 \right)^2 e^{4x}$

This is the derivative of the given expression using the product rule.

Would you like more details on any of these steps or explanations?

Related Questions:

1. What is the chain rule, and how does it apply to trigonometric functions?
2. How would you differentiate the expression if it were $e^{5x}$ instead of $e^{4x}$?
3. What other methods besides the product rule can be used to differentiate products of functions?
4. How does the differentiation of $\sin(2x)$ compare to that of $\sin(x)$?
5. Can the product rule be used for more than two functions? If so, how?

Tip:

When applying the product rule, it’s essential to break each part down and differentiate systematically to avoid confusion, especially when dealing with complex functions like products of trigonometric and exponential terms.