Let's go through the problems in the image one by one.

1. Find $\frac{dy}{dx}$

1. $y = \cos^3(4x)$

   • Differentiate using the chain rule: $\frac{dy}{dx} = 3\cos^2(4x) \cdot (-\sin(4x)) \cdot 4 = -12 \cos^2(4x) \sin(4x)$

2. $y = \left( \sin(x) + 1 \right)^4$

   • Apply the chain rule: $\frac{dy}{dx} = 4\left( \sin(x) + 1 \right)^3 \cdot \cos(x)$

3. $y = 7 \tan^3(1 - 2x)$

   • Differentiate using the chain rule: $\frac{dy}{dx} = 7 \cdot 3 \tan^2(1 - 2x) \cdot (-2) \sec^2(1 - 2x)$ Simplifying: $\frac{dy}{dx} = -42 \tan^2(1 - 2x) \sec^2(1 - 2x)$

4. $y = \sin^3(4x^3)$

   • Differentiate using the chain rule: $\frac{dy}{dx} = 3 \sin^2(4x^3) \cdot \cos(4x^3) \cdot 12x^2 = 36x^2 \sin^2(4x^3) \cos(4x^3)$

5. $y = \csc^2(3x)$

   • Use the chain rule: $\frac{dy}{dx} = 2 \csc^2(3x) \cdot (-\cot(3x)) \cdot 3 = -6 \csc^2(3x) \cot(3x)$

6. $y = (1 - \cos(3x))^3$

   • Use the chain rule: $\frac{dy}{dx} = 3(1 - \cos(3x))^2 \cdot \sin(3x) \cdot 3 = 9(1 - \cos(3x))^2 \sin(3x)$

7. $y = -3 \sec^3(2x)$

   • Differentiate using the chain rule: $\frac{dy}{dx} = -3 \cdot 3 \sec^2(2x) \cdot \sec(2x) \cdot \tan(2x) \cdot 2 = -18 \sec^3(2x) \tan(2x)$

8. $y = -2 \cos^5(3x)$

   • Use the chain rule: $\frac{dy}{dx} = -2 \cdot 5 \cos^4(3x) \cdot (-\sin(3x)) \cdot 3 = 30 \cos^4(3x) \sin(3x)$

9. $y = (1 - \tan^2(x))$

   • This simplifies using the identity $1 - \tan^2(x) = \sec^2(x)$, so: $\frac{dy}{dx} = 2 \sec^2(x) \cdot \sec^2(x) = 2 \sec^4(x)$

2. Find $\frac{dv}{dx}$

1. $v = \sin(3x) \cos(2x)$

   • Use the product rule and trigonometric identities: $\frac{dv}{dx} = \cos(3x) \cdot 3 \cos(2x) - \sin(3x) \cdot 2 \sin(2x)$

2. $v = \sin^3(2x + 1)$

   • Use the chain rule: $\frac{dv}{dx} = 3 \sin^2(2x + 1) \cdot \cos(2x + 1) \cdot 2 = 6 \sin^2(2x + 1) \cos(2x + 1)$

3. $v = 2x \cos(3x)$

   • Use the product rule: $\frac{dv}{dx} = 2 \cos(3x) - 2x \cdot 3 \sin(3x) = 2 \cos(3x) - 6x \sin(3x)$

4. $v = \sqrt{1 - \tan(3x)}$

   • Use the chain rule: $\frac{dv}{dx} = \frac{1}{2\sqrt{1 - \tan(3x)}} \cdot (-\sec^2(3x)) \cdot 3 = \frac{-3 \sec^2(3x)}{2\sqrt{1 - \tan(3x)}}$

5. $v = \sqrt{3 + 2 \cos(2x)}$

   • Use the chain rule: $\frac{dv}{dx} = \frac{1}{2\sqrt{3 + 2 \cos(2x)}} \cdot (-2 \sin(2x)) \cdot 2 = \frac{-2 \sin(2x)}{\sqrt{3 + 2 \cos(2x)}}$

Would you like further explanations on any of these derivatives? Here are 5 related questions for deeper understanding:

1. How can the chain rule be extended to multivariable functions?
2. What is the product rule and when is it applied?
3. How does the derivative of $\tan(x)$ result in $\sec^2(x)$?
4. Why is $\frac{d}{dx} \csc(x) = -\csc(x) \cot(x)$?
5. How do you apply trigonometric identities to simplify complex derivatives?

Tip: Always verify the derivative of trigonometric functions with identities before solving, as it often simplifies the work.