https://file.aiquickdraw.com/mathbot/file/1727609850130e4po1p9x.jpg

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.

1. Find dy/dx for the given functions. 2. Find dv/dx for the given functions.

Learn how to differentiate trigonometric functions using the chain and product rules. This guide includes detailed solutions for finding dy/dx and dv/dx for various functions involving trigonometric expressions. Suitable for students in Grades 11-12.

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