` If y=sin ^(2) 3x tan ^(3) 2x,then (dy)/(dx) =`

Options

A. ` sin 3x tan ^(2) 2x (2sin 3x sec ^(2) 2x +3cos 3x tan 3x) `
B. ` 2sin 3x tan ^(2) 2x (2sin 3x sec ^(2) 2x +3cos 3x tan 3x) `
C. ` 3sin 3x tan ^(2) 2x (2sin 3x sec ^(2) 2x +3cos 3x tan 3x) `
D. ` 6sin 3x tan ^(2) 2x (2sin 3x sec ^(2) 2x +3cos 3x tan 3x) `

Answer

To find the derivative of the function \( y = \sin^2(3x) \tan^3(2x) \), we will use the product rule of differentiation, which states that if \( y = u \cdot v \), then \( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = \sin^2(3x) \) and \( v = \tan^3(2x) \).

2. **Differentiate \( u \)**:
To differentiate \( u = \sin^2(3x) \), we will use the chain rule.
\[ \frac{du}{dx} = 2 \sin(3x) \cdot \cos(3x) \cdot \frac{d}{dx}(3x) = 2 \sin(3x) \cdot \cos(3x) \cdot 3 = 6 \sin(3x) \cos(3x) \]

3. **Differentiate \( v \)**:
To differentiate \( v = \tan^3(2x) \), we will also use the chain rule.
\[ \frac{dv}{dx} = 3 \tan^2(2x) \cdot \sec^2(2x) \cdot \frac{d}{dx}(2x) = 3 \tan^2(2x) \cdot \sec^2(2x) \cdot 2 = 6 \tan^2(2x) \sec^2(2x) \]

4. **Apply the product rule**:
Now, we can apply the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values we found:
\[ \frac{dy}{dx} = \sin^2(3x) \cdot (6 \tan^2(2x) \sec^2(2x)) + \tan^3(2x) \cdot (6 \sin(3x) \cos(3x)) \]

5. **Factor out common terms**:
We can factor out common terms from both parts:
\[ \frac{dy}{dx} = 6 \sin(3x) \cdot \sin(3x) \tan^2(2x) \sec^2(2x) + 6 \tan^3(2x) \sin(3x) \cos(3x) \]
\[ = 6 \sin(3x) \left( \sin^2(3x) \tan^2(2x) \sec^2(2x) + \tan^3(2x) \cos(3x) \right) \]

### Final Answer:
\[ \frac{dy}{dx} = 6 \sin(3x) \left( \sin^2(3x) \tan^2(2x) \sec^2(2x) + \tan^3(2x) \cos(3x) \right) \]