Let's find the derivative $\frac{dy}{dx}$ for the given function:

$y = e^{1 - x^2} \cdot \cos^4(2x + 3)$

Step-by-step solution:

1. Product Rule:
   Since $y = u \cdot v$, where $u = e^{1 - x^2}$ and $v = \cos^4(2x + 3)$, we use the product rule: $\frac{dy}{dx} = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}.$

2. Derivative of $u = e^{1 - x^2}$:
   The chain rule applies here. Let $z = 1 - x^2$, so: $\frac{du}{dx} = e^z \cdot \frac{dz}{dx}.$ Since $\frac{dz}{dx} = -2x$, we have: $\frac{du}{dx} = e^{1 - x^2} \cdot (-2x).$

3. Derivative of $v = \cos^4(2x + 3)$:
   This requires both the chain rule and the power rule. Let $w = \cos(2x + 3)$, so: $\frac{dv}{dx} = 4 \cdot w^3 \cdot \frac{dw}{dx}.$ Now, $\frac{dw}{dx} = -\sin(2x + 3) \cdot \frac{d}{dx}(2x + 3) = -\sin(2x + 3) \cdot 2$. Substituting: $\frac{dv}{dx} = 4 \cdot \cos^3(2x + 3) \cdot (-\sin(2x + 3)) \cdot 2.$ Simplify: $\frac{dv}{dx} = -8 \cos^3(2x + 3) \sin(2x + 3).$

4. Combine using the product rule: Substitute $\frac{du}{dx}$ and $\frac{dv}{dx}$ into the product rule: $\frac{dy}{dx} = \big(e^{1 - x^2} \cdot (-2x)\big) \cdot \cos^4(2x + 3) + e^{1 - x^2} \cdot \big(-8 \cos^3(2x + 3) \sin(2x + 3)\big).$

5. Simplify the expression: $\frac{dy}{dx} = -2x e^{1 - x^2} \cos^4(2x + 3) - 8 e^{1 - x^2} \cos^3(2x + 3) \sin(2x + 3).$

Final Answer:

$\frac{dy}{dx} = -2x e^{1 - x^2} \cos^4(2x + 3) - 8 e^{1 - x^2} \cos^3(2x + 3) \sin(2x + 3).$

Would you like a detailed explanation of any specific part?

Related Questions:

1. How do you apply the product rule in similar problems?
2. Can we simplify the trigonometric term $\cos^4$ further using identities?
3. How does the chain rule work for exponential functions like $e^{f(x)}$?
4. What are some strategies to simplify derivatives with trigonometric powers?
5. Can you explain how $\sin(2x + 3)$ and $\cos(2x + 3)$ behave graphically in this problem?

Tip:

When differentiating composite functions, always identify the "outer" and "inner" layers and use the chain rule systematically to avoid errors.